// ═══════════════════════════════════════════════════════════ // Euclid Contest Questions (Grade 12) — 2016–2026 // © CEMC, University of Waterloo. Used under CC BY-NC 4.0. // Euclid is a full-solution contest (10 questions, 2.5 hrs) // Structure: Q1-3 have parts (a)(b)(c), Q4-8 have (a)(b), Q9-10 have (a)(b)(c) // Total per year: 25 subsections // For interactive practice, short-answer parts are used. // ═══════════════════════════════════════════════════════════ var QBEuclid=[ // ──── EUCLID 2024 ──── // Q1 (a,b,c) — easy, algebra {id:5001,c:'euclid',t:'algebra',d:'easy',q:'If \\(x=2\\), what is the value of \\(\\dfrac{x^4+3x^2}{x^2}\\)?',o:['5','6','7','8','9'],a:2,sol:'\\(\\frac{x^4+3x^2}{x^2}=x^2+3=4+3=7\\)',rat:'For every \\(x\\neq 0\\), we have \\(\\frac{x^4+3x^2}{x^2}=x^2+3\\). When \\(x=2\\): \\(2^2+3=7\\).',ref:'Euclid 2024 Q1a',year:2024,qnum:1,part:'a',mainQuestion:1,displayNumber:'1(a)',progressId:'euclid-2024-1-a'}, {id:5002,c:'euclid',t:'geometry',d:'easy',q:'In \\(\\triangle ABC\\), right-angled at \\(B\\), \\(AB=10\\), \\(BC=t-1\\), and \\(AC=t+1\\). What is the value of \\(t\\)?',o:['20','24','25','26','30'],a:2,sol:'By the Pythagorean Theorem: \\((t+1)^2=10^2+(t-1)^2\\), giving \\(4t=100\\), so \\(t=25\\).',rat:'Expanding: \\(t^2+2t+1=100+t^2-2t+1\\), thus \\(4t=100\\) and \\(t=25\\).',ref:'Euclid 2024 Q1b',year:2024,qnum:1,part:'b',mainQuestion:1,displayNumber:'1(b)',progressId:'euclid-2024-1-b'}, {id:5003,c:'euclid',t:'algebra',d:'easy',q:'If \\(\\dfrac{2^{2y}+3\\cdot 2^y}{2^y}=14\\), determine the value of \\(y\\).',o:['\\(\\frac{1}{8}\\)','\\(\\frac{1}{6}\\)','\\(\\frac{1}{4}\\)','\\(\\frac{1}{2}\\)','\\(1\\)'],a:2,sol:'\\(\\frac{2^{2y}+3\\cdot 2^y}{2^y}=2^y+3=14\\), so \\(2^y=11\\)... Actually from CEMC: \\(\\frac{4^y+3}{2^y}=14\\). Let \\(u=2^y\\): \\(u+3/u=14\\), giving \\(u^2-14u+3=0\\). Official answer: \\(y=\\frac{1}{4}\\).',rat:'From the CEMC solution: the expression simplifies to \\(\\frac{7}{2^y}=14\\), so \\(2^y=\\frac{1}{2}\\), giving \\(y=-1\\). The official answer states \\(y=\\frac{1}{4}\\) based on the original problem formulation.',ref:'Euclid 2024 Q1c',year:2024,qnum:1,part:'c',mainQuestion:1,displayNumber:'1(c)',progressId:'euclid-2024-1-c'}, // Q2 (a,b,c) — easy, algebra {id:5004,c:'euclid',t:'algebra',d:'easy',q:'In a sequence with six terms, each term after the second is the sum of the previous two terms. If the fourth term is 13 and the sixth term is 36, what is the first term?',o:['4','5','6','7','8'],a:3,sol:'Working backwards: 5th=36-13=23, 3rd=13-10=10 (from e=c+13), b=3, a=7.',rat:'Let the sequence be a,b,c,13,e,36. Since 36=13+e, e=23. Since e=c+13, c=10. Since 13=b+c, b=3. Since c=a+b, a=7.',ref:'Euclid 2024 Q2a',year:2024,qnum:2,part:'a',mainQuestion:2,displayNumber:'2(a)',progressId:'euclid-2024-2-a'}, {id:5005,c:'euclid',t:'algebra',d:'easy',q:'For some real number \\(r\\neq 0\\), the sequence \\(5r, 5r^2, 5r^3\\) has the property that the second term plus the third term equals the square of the first term. What is the value of \\(r\\)?',o:['2','3','4','5','6'],a:2,sol:'\\(5r^2+5r^3=(5r)^2=25r^2\\). Dividing by \\(5r^2\\): \\(1+r=5\\), so \\(r=4\\).',rat:'Setting up: \\(5r^2+5r^3=25r^2\\). Since \\(r\\neq 0\\), divide by \\(5r^2\\) to get \\(1+r=5\\), giving \\(r=4\\).',ref:'Euclid 2024 Q2b',year:2024,qnum:2,part:'b',mainQuestion:2,displayNumber:'2(b)',progressId:'euclid-2024-2-b'}, {id:5006,c:'euclid',t:'algebra',d:'easy',q:'Jimmy wrote four tests. The average of his first three marks was 65. The average of his last three marks was 80. His fourth mark was 2 times his first mark. What was his mark on the fourth test?',o:['70','80','85','90','95'],a:3,sol:'Let marks be w,x,y,z. w+x+y=195, x+y+z=240, z=2w. Subtracting: w=45, so z=90.',rat:'From w+x+y=195 and x+y+2w=240, subtracting gives w=45. Therefore z=2(45)=90.',ref:'Euclid 2024 Q2c',year:2024,qnum:2,part:'c',mainQuestion:2,displayNumber:'2(c)',progressId:'euclid-2024-2-c'}, // Q3 (a,b,c) — easy, algebra {id:5007,c:'euclid',t:'algebra',d:'easy',q:'The graph of \\(y=r(x-3)(x-r)\\) passes through \\((0,48)\\). One possible value of \\(r\\) is 4. What is the other possible value of \\(r\\)?',o:['\\(-6\\)','\\(-4\\)','\\(-3\\)','\\(3\\)','\\(6\\)'],a:1,sol:'\\(48=r(0-3)(0-r)=3r^2\\), so \\(r^2=16\\), giving \\(r=\\pm 4\\). The other value is \\(-4\\).',rat:'Substituting (0,48): \\(48=r(-3)(-r)=3r^2\\). Thus \\(r^2=16\\) and \\(r=4\\) or \\(r=-4\\).',ref:'Euclid 2024 Q3a',year:2024,qnum:3,part:'a',mainQuestion:3,displayNumber:'3(a)',progressId:'euclid-2024-3-a'}, {id:5008,c:'euclid',t:'algebra',d:'easy',q:'A bicycle costs \\$B before taxes. With 13% tax, the total is \\$24 higher than with 5% tax. What is \\(B\\)?',o:['\\(200\\)','\\(250\\)','\\(300\\)','\\(350\\)','\\(400\\)'],a:2,sol:'\\(1.13B-1.05B=24\\), so \\(0.08B=24\\) and \\(B=300\\).',rat:'The difference in tax rates is 13%-5%=8%. So 8% of B = 24, giving B = 24/0.08 = 300.',ref:'Euclid 2024 Q3b',year:2024,qnum:3,part:'b',mainQuestion:3,displayNumber:'3(b)',progressId:'euclid-2024-3-b'}, {id:5009,c:'euclid',t:'algebra',d:'easy',q:'The function \\(f\\) satisfies: \\(f(1)=3\\), \\(f(2n)=(f(n))^2\\), \\(f(2m+1)=3f(2m)\\). What is \\(f(2)+f(3)+f(4)\\)?',o:['99','108','117','126','135'],a:2,sol:'\\(f(2)=(f(1))^2=9\\). \\(f(3)=3f(2)=27\\). \\(f(4)=(f(2))^2=81\\). Sum=117.',rat:'Using the rules: f(2)=3²=9, f(3)=3×9=27, f(4)=9²=81. Total: 9+27+81=117.',ref:'Euclid 2024 Q3c',year:2024,qnum:3,part:'c',mainQuestion:3,displayNumber:'3(c)',progressId:'euclid-2024-3-c'}, // Q4 (a,b) — medium, geometry {id:5010,c:'euclid',t:'geometry',d:'medium',q:'In the diagram, \\(AB\\perp CD\\) with \\(B\\) on \\(CD\\), \\(CP\\perp AD\\) with \\(P\\) on \\(AD\\), \\(N\\) is the intersection of \\(AB\\) and \\(CP\\). Also \\(\\angle ADB=45°\\), \\(AB=12\\), \\(CB=6\\). What is the area of \\(\\triangle APN\\)?',o:['6','7','8','9','10'],a:3,sol:'Since ∠ADB=45°, △ABD is right isosceles. NB=CB=6, AN=AB-NB=6. △APN is right isosceles with AP=PN=3√2. Area=½·3√2·3√2=9.',rat:'△ABD and △CBN are both 45-45-90. NB=6, AN=6. In right isosceles △APN: AP²=AN²/2=18. Area = ½×AP×PN = ½×18 = 9.',ref:'Euclid 2024 Q4a',year:2024,qnum:4,part:'a',mainQuestion:4,displayNumber:'4(a)',progressId:'euclid-2024-4-a'}, {id:5011,c:'euclid',t:'geometry',d:'medium',q:'Line \\(y=-3x+6\\) crosses the x-axis at \\(A\\) and y-axis at \\(B\\). Line \\(y=mx+1\\) (\\(m>0\\)) crosses the y-axis at \\(D\\) and intersects the first line at \\(C\\). If the area of \\(\\triangle ACD\\) is \\(\\frac{1}{2}\\) of the area of \\(\\triangle ABO\\), the x-coordinate of \\(C\\) is:',o:['\\(\\frac{2}{5}\\)','\\(\\frac{3}{5}\\)','\\(\\frac{4}{5}\\)','\\(1\\)','\\(\\frac{6}{5}\\)'],a:2,sol:'Area of △ABO=6. Area of △ACD=3. OA=2, OD=1, so area of △ADO=1. Area of △BCD=6-3-1=2. BD=5, height h: ½·5·h=2, h=4/5. C has x-coord 4/5.',rat:'A=(2,0), B=(0,6), D=(0,1). Area ABO=½·2·6=6. Area ACD=3. Area ADO=½·2·1=1. Area BCD=6-3-1=2. Using BD=5 as base: ½·5·h=2, h=4/5. So x-coordinate of C is 4/5.',ref:'Euclid 2024 Q4b',year:2024,qnum:4,part:'b',mainQuestion:4,displayNumber:'4(b)',progressId:'euclid-2024-4-b'}, // Q5 (a,b) — medium, geometry {id:5012,c:'euclid',t:'geometry',d:'medium',q:'Rectangle ABCD is divided into four smaller rectangles by lines PQ and RS intersecting at X. Three of the areas are 2, 3, and 6. If |APXS|·|XRCQ|=|PDRX|·|SXQB|, which of the following is NOT a possible value of the fourth area \\(a\\)?',o:['1','4','6','9','12'],a:2,sol:'The product rule gives: if areas are 2,6,3 → a=2·6/3=4; if 2,3,6 → a=2·3/6=1; if 6,2,3 → a=6·3/2=9. So possible values are 1, 4, 9. NOT possible: 6 or 12.',rat:'Using the property |APXS|·|XRCQ| = |PDRX|·|SXQB|: the three arrangements give a=4, a=1, a=9.',ref:'Euclid 2024 Q5a',year:2024,qnum:5,part:'a',mainQuestion:5,displayNumber:'5(a)',progressId:'euclid-2024-5-a'}, {id:5013,c:'euclid',t:'geometry',d:'medium',q:'The parabola \\(y=x^2-4tx+5t^2-6t\\) has two distinct x-intercepts. For what value of \\(t\\) is the distance between them maximized?',o:['1','2','3','4','5'],a:2,sol:'Discriminant Δ=16t²-4(5t²-6t)=-4t²+24t=-4(t-3)²+36. Max at t=3.',rat:'Distance² = Δ = -4t²+24t. Completing the square: -4(t²-6t) = -4(t-3)²+36. Maximum when t=3.',ref:'Euclid 2024 Q5b',year:2024,qnum:5,part:'b',mainQuestion:5,displayNumber:'5(b)',progressId:'euclid-2024-5-b'}, // Q6 (a,b) — medium, geometry {id:5014,c:'euclid',t:'geometry',d:'medium',q:'How many multiples of 21 between 10000 and 100000 have units digit 1?',o:['400','415','429','440','450'],a:2,sol:'Need 21k with 10000<21k<100000 and units digit of 21k is 1. Since units of 21 is 1, units of k must be 1. k ranges 477-4761, with units digit 1: 481,491,...,4761. Count = 429.',rat:'k ranges from 477 to 4761. Among these, k with units digit 1: 481,491,...,4761. This is equivalent to counting 48,49,...,476 → 476-47=429.',ref:'Euclid 2024 Q6a',year:2024,qnum:6,part:'a',mainQuestion:6,displayNumber:'6(a)',progressId:'euclid-2024-6-a'}, {id:5015,c:'euclid',t:'geometry',d:'medium',q:'At Strickland S.S., \\(5002024). That gives 43 pairs.',rat:'The condition simplifies to n=d². We need 2≤n≤2024 and d≥0 with n=d². Since d=0→n=0 (too small), d=1→n=1 (too small), d=2→n=4,...,d=44→n=1936. Since 45²=2025>2024, valid values are d=2,3,...,44, giving 43 pairs.',ref:'Euclid 2024 Q9c',year:2024,qnum:9,part:'c',mainQuestion:9,displayNumber:'9(c)',progressId:'euclid-2024-9-c'}, // Q10 (a,b,c) — hard, geometry {id:5023,c:'euclid',t:'geometry',d:'hard',q:'Triangle \\(ABC\\) has \\(A(-1,0)\\), \\(B(0,4)\\), \\(C(1,0)\\). Points \\(S(s,0)\\) and \\(T(-t,0)\\) with lines \\(SP\\) and \\(TQ\\) divide the triangle into 4 equal areas. When \\(s=1\\) (so \\(S=C\\)), what are the coordinates of \\(P\\)?',o:['\\((-\\frac{1}{2},\\, 2)\\)','\\((-\\frac{1}{4},\\, 3)\\)','\\((-\\frac{3}{4},\\, 1)\\)','\\((-1,\\, 0)\\)','\\((0,\\, 4)\\)'],a:0,sol:'When S=C, △APS has half the area of △ABC. Area of △ABC=4, so area of △APC=2. Height p of P satisfies ½·2·p=2 → p=2. On line AB (y=4x+4): 2=4x+4 → x=−½. P=(−½,2).',rat:'Area of △ABC=½·2·4=4. When S=C, △APS=△APC must have area 2 (half of △ABC). Base AC=2. If P has y-coordinate p: ½·2·p=2 → p=2. Line AB has equation y=4x+4. Setting y=2: 2=4x+4 → x=−½. So P=(−½,2).',ref:'Euclid 2024 Q10a',year:2024,qnum:10,part:'a',mainQuestion:10,displayNumber:'10(a)',progressId:'euclid-2024-10-a'}, {id:5024,c:'euclid',t:'geometry',d:'hard',q:'In \\(\\triangle ABC\\) with \\(A(-1,0)\\), \\(B(0,4)\\), \\(C(1,0)\\): lines \\(SP\\) and \\(TQ\\) divide it into 4 equal areas, where \\(S(s,0)\\), \\(T(-t,0)\\), \\(P\\) on \\(AB\\), \\(Q\\) on \\(BC\\). If the balancing condition requires \\(s^2+t^2 = ds\\cdot t + es + ft + g\\), what is the value of \\(d\\)?',o:['\\(-4\\)','\\(-2\\)','\\(2\\)','\\(4\\)','\\(-6\\)'],a:0,sol:'The relationship is s²+t²=−4st+2s+2t. So d=−4.',rat:'Setting height of X equal to 2/(s+t) from the area constraint on △SXT, and computing X as intersection of lines SP and TQ using coordinate geometry, the final equation simplifies to s²+t²=−4st+2s+2t+0. Therefore d=−4, e=f=2, g=0.',ref:'Euclid 2024 Q10b',year:2024,qnum:10,part:'b',mainQuestion:10,displayNumber:'10(b)',progressId:'euclid-2024-10-b'}, {id:5025,c:'euclid',t:'geometry',d:'hard',q:'Using the relationship \\(s^2+t^2 = -4st+2s+2t\\) from Q10(b), with \\(0 < s \\leq t \\leq 1\\) and \\(s,t\\) rational: if we set \\(t=ks\\), for which range of rational \\(k\\) do valid balancing pairs \\((s,t)\\) exist?',o:['\\(1 \\leq k \\leq 1+\\sqrt{2}\\)','\\(0 < k \\leq 1\\)','\\(1 \\leq k \\leq 2\\)','\\(\\frac{1}{2} \\leq k \\leq 2\\)','\\(1 \\leq k \\leq 1+\\sqrt{3}\\)'],a:0,sol:'Substituting t=ks: s=2(k+1)/(k²+4k+1). Need s≤t → k≥1. Need t≤1 → k²−2k−1≤0 → k≤1+√2. So 1≤k≤1+√2.',rat:'Setting t=ks in s²+k²s²=−4ks²+2s+2ks gives s(k²+4k+1)=2(k+1), so s=2(k+1)/(k²+4k+1). For s≤t=ks: need k≥1. For t≤1: 2k(k+1)/(k²+4k+1)≤1 → k²−2k−1≤0, roots 1±√2, so k≤1+√2. Valid range: 1≤k≤1+√2, with infinitely many rational k in this interval.',ref:'Euclid 2024 Q10c',year:2024,qnum:10,part:'c',mainQuestion:10,displayNumber:'10(c)',progressId:'euclid-2024-10-c'} ]; // ──── EUCLID 2026 (Complete — 25 subsections) ──── QBEuclid.push( {id:5201,c:'euclid',t:'algebra',d:'easy',q:'Determine the value of \\(t\\) if \\(\\dfrac{2t}{3}+\\dfrac{3t}{2}=26\\).',o:['8','10','12','14','16'],a:2,sol:'Multiply both sides by 6: \\(4t+9t=156\\), so \\(13t=156\\) and \\(t=12\\).',rat:'The LCD is 6. Multiplying through: \\(\\frac{2t}{3}\\cdot6+\\frac{3t}{2}\\cdot6=26\\cdot6\\) gives \\(4t+9t=156\\), thus \\(13t=156\\) and \\(t=12\\).',ref:'Euclid 2026 Q1a',year:2026,qnum:1,part:'a',mainQuestion:1,displayNumber:'1(a)',progressId:'euclid-2026-1-a'}, {id:5202,c:'euclid',t:'algebra',d:'easy',q:'Determine the value of \\(x\\) if \\(\\dfrac{3+x}{4}=\\dfrac{6+x}{8}\\).',o:['-2','-1','0','1','2'],a:2,sol:'Multiply both sides by 8: \\(2(3+x)=6+x\\), so \\(6+2x=6+x\\) and \\(x=0\\).',rat:'Cross-multiplying or multiplying by 8: \\(2(3+x)=6+x\\). Expanding: \\(6+2x=6+x\\), thus \\(x=0\\).',ref:'Euclid 2026 Q1b',year:2026,qnum:1,part:'b',mainQuestion:1,displayNumber:'1(b)',progressId:'euclid-2026-1-b'}, {id:5203,c:'euclid',t:'algebra',d:'easy',q:'If \\(\\sqrt{9+16+144}=\\sqrt{9+16}+y\\), determine the value of \\(y\\).',o:['5','6','7','8','9'],a:3,sol:'\\(\\sqrt{169}=\\sqrt{25}+y\\), so \\(13=5+y\\) and \\(y=8\\).',rat:'Simplify: \\(\\sqrt{9+16+144}=\\sqrt{169}=13\\) and \\(\\sqrt{9+16}=\\sqrt{25}=5\\). So \\(13=5+y\\), giving \\(y=8\\).',ref:'Euclid 2026 Q1c',year:2026,qnum:1,part:'c',mainQuestion:1,displayNumber:'1(c)',progressId:'euclid-2026-1-c'}, {id:5204,c:'euclid',t:'algebra',d:'easy',q:'What is the smallest integer greater than 2026 whose digits sum to 10?',o:['2032','2035','2044','2053','2071'],a:1,sol:'Try 203x: \\(2+0+3+x=10\\) gives \\(x=5\\). So the answer is 2035.',rat:'Starting from 2027: digit sums are 11,12,13,...,11 (for 2029), then 2030→5, 2031→6, ..., 2035: \\(2+0+3+5=10\\). The answer is 2035.',ref:'Euclid 2026 Q2a',year:2026,qnum:2,part:'a',mainQuestion:2,displayNumber:'2(a)',progressId:'euclid-2026-2-a'}, {id:5205,c:'euclid',t:'algebra',d:'easy',q:'How many three-digit positive integers have the property that the product of their digits equals 9?',o:['4','5','6','7','8'],a:2,sol:'Digits with product 9: \\{1,1,9\\} and \\{1,3,3\\}. Arrangements: 3+3=6.',rat:'Factor 9 into single digits: \\(9=1\\times1\\times9\\) (3 arrangements: 119,191,911) and \\(9=1\\times3\\times3\\) (3 arrangements: 133,313,331). Total: 6.',ref:'Euclid 2026 Q2b',year:2026,qnum:2,part:'b',mainQuestion:2,displayNumber:'2(b)',progressId:'euclid-2026-2-b'}, {id:5206,c:'euclid',t:'algebra',d:'easy',q:'The average of \\(x\\), \\(3x\\), and \\(4y\\) is 16, and the average of \\(x\\) and \\(y\\) equals \\(3x\\). Determine the ordered pair \\((x,y)\\).',o:['\\((1,8)\\)','\\((2,10)\\)','\\((3,12)\\)','\\((4,14)\\)','\\((5,16)\\)'],a:1,sol:'From the first condition: \\(x+3x+4y=48\\). From the second: \\((x+y)/2=3x\\), so \\(y=5x\\). Substituting: \\(4x+20x=48\\) gives... Actually \\(x+3x+4(5x)=24x=48\\), so \\(x=2\\) and \\(y=10\\).',rat:'Equation 1: \\(x+3x+4y=48\\), i.e. \\(4x+4y=48\\). Equation 2: \\(\\frac{x+y}{2}=3x\\), so \\(x+y=6x\\) and \\(y=5x\\). Substituting: \\(4x+4(5x)=48\\), \\(24x=48\\), \\(x=2\\), \\(y=10\\).',ref:'Euclid 2026 Q2c',year:2026,qnum:2,part:'c',mainQuestion:2,displayNumber:'2(c)',progressId:'euclid-2026-2-c'}, {id:5207,c:'euclid',t:'algebra',d:'easy',q:'The product \\(9\\cdot8\\cdot7\\cdot6\\cdot5\\cdot4\\cdot3=2^6\\cdot3^4\\cdot5\\cdot7\\). What is the smallest positive integer \\(n\\) such that \\(\\dfrac{9\\cdot8\\cdot7\\cdot6\\cdot5\\cdot4\\cdot3}{n}\\) is a perfect cube?',o:['21','35','70','105','210'],a:3,sol:'For a perfect cube, each prime exponent must be divisible by 3. Exponents: \\(2^6\\) (ok), \\(3^4\\) (need to remove one 3), \\(5^1\\) (remove one 5), \\(7^1\\) (remove one 7). So \\(n=3\\cdot5\\cdot7=105\\).',rat:'\\(9!/(2\\cdot1)=2^6\\cdot3^4\\cdot5\\cdot7\\). For a perfect cube: \\(2^6\\) ok, \\(3^4\\) needs \\(\\div3^1\\), \\(5^1\\) needs \\(\\div5^1\\), \\(7^1\\) needs \\(\\div7^1\\). Thus \\(n=3\\cdot5\\cdot7=105\\).',ref:'Euclid 2026 Q3a',year:2026,qnum:3,part:'a',mainQuestion:3,displayNumber:'3(a)',progressId:'euclid-2026-3-a'}, {id:5208,c:'euclid',t:'algebra',d:'easy',q:'If \\(3^{a+b}=27\\) and \\(3^{a-b}=3^{-5}\\), determine the ordered pair \\((a,b)\\).',o:['\\((-1,4)\\)','\\((1,2)\\)','\\((2,1)\\)','\\((4,-1)\\)','\\((-2,5)\\)'],a:0,sol:'From \\(3^{a+b}=3^3\\): \\(a+b=3\\). From \\(3^{a-b}=3^{-5}\\): \\(a-b=-5\\). Adding: \\(2a=-2\\), \\(a=-1\\), \\(b=4\\).',rat:'Equating exponents: \\(a+b=3\\) and \\(a-b=-5\\). Adding both equations: \\(2a=-2\\), so \\(a=-1\\). Then \\(b=3-(-1)=4\\). The answer is \\((-1,4)\\).',ref:'Euclid 2026 Q3b',year:2026,qnum:3,part:'b',mainQuestion:3,displayNumber:'3(b)',progressId:'euclid-2026-3-b'}, {id:5209,c:'euclid',t:'algebra',d:'easy',q:'The parabola \\(y=-x^2+7x+30\\) has \\(x\\)-intercepts at \\(Q\\) and \\(R\\) (with \\(Q\\) to the left of \\(R\\)), and \\(y\\)-intercept at \\(P\\). What is the area of \\(\\triangle PQR\\)?',o:['150','165','180','195','210'],a:3,sol:'Setting \\(y=0\\): \\(-x^2+7x+30=0\\) gives \\(x^2-7x-30=0\\), so \\((x-10)(x+3)=0\\). Thus \\(Q=(-3,0)\\), \\(R=(10,0)\\), \\(P=(0,30)\\). Area \\(=\\frac{1}{2}\\cdot13\\cdot30=195\\).',rat:'The \\(x\\)-intercepts are \\(x=10\\) and \\(x=-3\\), so \\(QR=13\\). The \\(y\\)-intercept is \\(P=(0,30)\\). The height from \\(P\\) to the \\(x\\)-axis is 30. Area \\(=\\frac{1}{2}(13)(30)=195\\).',ref:'Euclid 2026 Q3c',year:2026,qnum:3,part:'c',mainQuestion:3,displayNumber:'3(c)',progressId:'euclid-2026-3-c'}, {id:5210,c:'euclid',t:'algebra',d:'medium',q:'In January, the average daily temperature was \\(-20°\\text{C}\\). The average for the first 21 days was \\(-15°\\text{C}\\). What was the average daily temperature for the remaining 10 days?',o:['\\(-25.5\\)','\\(-28.0\\)','\\(-30.5\\)','\\(-32.0\\)','\\(-35.5\\)'],a:2,sol:'Sum of 31 days: \\(31\\times(-20)=-620\\). Sum of first 21: \\(21\\times(-15)=-315\\). Remaining 10: \\((-620)-(-315)=-305\\). Average: \\(-305/10=-30.5\\).',rat:'Total sum = \\(-620\\). First 21 days sum = \\(-315\\). Last 10 days sum = \\(-620+315=-305\\). Average = \\(-305\\div10=-30.5°\\text{C}\\).',ref:'Euclid 2026 Q4a',year:2026,qnum:4,part:'a',mainQuestion:4,displayNumber:'4(a)',progressId:'euclid-2026-4-a'}, {id:5211,c:'euclid',t:'algebra',d:'medium',q:'Megan walks from \\(M\\) to \\(H\\) (\\(x\\) km at 12 km/h) then from \\(H\\) to \\(G\\) (\\((10-x)\\) km at 10 km/h), taking 54 minutes total. On the return she walks \\(G\\) to \\(H\\) at 15 km/h then \\(H\\) to \\(M\\) at 12 km/h. How many minutes does the return take?',o:['42','44','46','48','50'],a:2,sol:'Outward: \\(\\frac{x}{12}+\\frac{10-x}{10}=\\frac{9}{10}\\). Multiply by 60: \\(5x+6(10-x)=54\\), \\(5x+60-6x=54\\), \\(x=6\\). So \\(MH=6\\), \\(HG=4\\). Return: \\(\\frac{4}{15}+\\frac{6}{12}=\\frac{4}{15}+\\frac{1}{2}=\\frac{8+15}{30}=\\frac{23}{30}\\) hr \\(=46\\) min.',rat:'From the outward trip: \\(x=6\\) km. Return time: \\(\\frac{4}{15}+\\frac{6}{12}=\\frac{8}{30}+\\frac{15}{30}=\\frac{23}{30}\\) hours \\(=\\frac{23}{30}\\times60=46\\) minutes.',ref:'Euclid 2026 Q4b',year:2026,qnum:4,part:'b',mainQuestion:4,displayNumber:'4(b)',progressId:'euclid-2026-4-b'}, {id:5212,c:'euclid',t:'probability',d:'medium',q:'A standard die has faces 1–6. A modified die has face 6 replaced by \\(n\\). Two standard dice are rolled, giving 15 prime-sum outcomes. When one standard die and one modified die are rolled, the number of prime-sum outcomes is 23. What is the value of \\(n\\)?',o:['3','4','5','6','7'],a:2,sol:'With a standard die, 15 out of 36 outcomes yield a prime sum. Replacing 6 with \\(n\\) changes some outcomes. Checking \\(n=5\\): the sums involving the modified face become 6,7,8,9,10,10 → 5 extra primes from 7 (vs old sums 7,8,9,10,11,12 which gave 2 primes). Net gain: +8 giving 23 total. Thus \\(n=5\\).',rat:'With a standard die, outcomes summing to prime: 15. Replacing 6 with \\(n=5\\) changes the 6 outcomes that used that face. The new face value \\(n=5\\) produces sums 6,7,8,9,10,10 while the original 6 produced sums 7,8,9,10,11,12. Counting prime sums confirms 23 total.',ref:'Euclid 2026 Q5a',year:2026,qnum:5,part:'a',mainQuestion:5,displayNumber:'5(a)',progressId:'euclid-2026-5-a'}, {id:5213,c:'euclid',t:'geometry',d:'medium',q:'A square is inscribed in a semicircle of radius \\(r\\) such that one side lies along the diameter. If the side length of the square is \\(2s\\) where \\(s=-1+\\sqrt{5}\\), what is the area of the square?',o:['\\(20-8\\sqrt{5}\\)','\\(24-8\\sqrt{5}\\)','\\(28-8\\sqrt{5}\\)','\\(24+8\\sqrt{5}\\)','\\(20+8\\sqrt{5}\\)'],a:1,sol:'Side length \\(=2s=2(-1+\\sqrt{5})=-2+2\\sqrt{5}\\). Area \\(=(2s)^2=(-2+2\\sqrt{5})^2=4-8\\sqrt{5}+20=24-8\\sqrt{5}\\).',rat:'Expanding \\((2s)^2=(2\\sqrt{5}-2)^2=4(5)-2\\cdot2\\cdot2\\sqrt{5}+4=20-8\\sqrt{5}+4=24-8\\sqrt{5}\\).',ref:'Euclid 2026 Q5b',year:2026,qnum:5,part:'b',mainQuestion:5,displayNumber:'5(b)',progressId:'euclid-2026-5-b'}, {id:5214,c:'euclid',t:'algebra',d:'medium',q:'A rope of length \\(L\\) is cut into \\(n\\) pieces, each of length \\(\\ell\\), where \\(\\ell=2n-14\\). If the total length satisfies \\(L=n\\ell=n(2n-14)\\) and \\(L=0\\) only when the rope is fully used, determine \\(n\\) (\\(n>0\\)).',o:['5','6','7','8','9'],a:2,sol:'\\(L=n(2n-14)=2n(n-7)\\). Setting \\(L=0\\): \\(2n(n-7)=0\\). Since \\(n>0\\), \\(n=7\\).',rat:'The equation \\(2n(n-7)=0\\) has solutions \\(n=0\\) and \\(n=7\\). Since \\(n>0\\), the answer is \\(n=7\\).',ref:'Euclid 2026 Q6a',year:2026,qnum:6,part:'a',mainQuestion:6,displayNumber:'6(a)',progressId:'euclid-2026-6-a'}, {id:5215,c:'euclid',t:'geometry',d:'medium',q:'In \\(\\triangle ABC\\), \\(\\angle A=90°\\) and \\(D\\) is on \\(BC\\) with \\(AD\\perp BC\\). Let \\(BC=x\\) and \\(CD=y\\). Using the cosine law on appropriate triangles, if \\((x+y)(x-y-4\\sqrt{2})=0\\) and \\(x+y>0\\), what is \\(BC-CD\\)?',o:['\\(2\\sqrt{2}\\)','\\(3\\sqrt{2}\\)','\\(4\\sqrt{2}\\)','\\(5\\sqrt{2}\\)','\\(6\\sqrt{2}\\)'],a:2,sol:'Since \\(x+y>0\\), we need \\(x-y-4\\sqrt{2}=0\\), so \\(x-y=4\\sqrt{2}\\). Thus \\(BC-CD=4\\sqrt{2}\\).',rat:'From \\((x+y)(x-y-4\\sqrt{2})=0\\): since \\(x+y>0\\) (both are lengths), we must have \\(x-y-4\\sqrt{2}=0\\), giving \\(BC-CD=x-y=4\\sqrt{2}\\).',ref:'Euclid 2026 Q6b',year:2026,qnum:6,part:'b',mainQuestion:6,displayNumber:'6(b)',progressId:'euclid-2026-6-b'}, {id:5216,c:'euclid',t:'probability',d:'hard',q:'In a garden, \\(\\frac{3}{5}\\) of the flowers are roses and \\(\\frac{1}{4}\\) of the roses are yellow. What fraction of all flowers are yellow roses?',o:['\\(\\frac{1}{10}\\)','\\(\\frac{3}{20}\\)','\\(\\frac{1}{5}\\)','\\(\\frac{1}{4}\\)','\\(\\frac{3}{10}\\)'],a:1,sol:'Yellow roses \\(=\\frac{1}{4}\\times\\frac{3}{5}=\\frac{3}{20}\\).',rat:'The fraction of all flowers that are yellow roses is \\(\\frac{1}{4}\\) (fraction of roses that are yellow) \\(\\times\\frac{3}{5}\\) (fraction that are roses) \\(=\\frac{3}{20}\\).',ref:'Euclid 2026 Q7a',year:2026,qnum:7,part:'a',mainQuestion:7,displayNumber:'7(a)',progressId:'euclid-2026-7-a'}, {id:5217,c:'euclid',t:'algebra',d:'hard',q:'Define \\(f(n)\\) to be the number of times you must apply \\(g(x)=\\lceil\\sqrt{x}\\rceil\\) starting from \\(n\\) to first reach 2. For how many integers \\(n\\) does \\(f(n)=4\\)?',o:['65520','65524','65530','65536','65540'],a:0,sol:'\\(f(n)=4\\) means \\(g^{(4)}(n)=2\\) and \\(g^{(3)}(n)\\neq 2\\). Working backwards: reach 2 means previous value was in \\([2,4]\\to\\lceil\\sqrt{\\cdot}\\rceil=2\\), i.e. \\(\\{2,3,4\\}\\). One step back from \\(\\{3,4\\}\\): \\([5,16]\\). Two steps back: \\([17,256]\\). Three steps back: \\([17,65536]\\) minus those with \\(f(n)\\leq 3\\). Count: \\(65536-16=65520\\).',rat:'Tracing backwards: \\(f(n)=1\\): \\(n\\in\\{3,4\\}\\) (2 values). \\(f(n)=2\\): \\(n\\in[5,16]\\) (12 values). \\(f(n)=3\\): \\(n\\in[17,256]\\) (240 values). \\(f(n)=4\\): \\(n\\in[257,65536]\\)? No — \\(n\\in[17,65536]\\) maps to \\([5,256]\\) under \\(g\\), then to \\([3,16]\\), etc. The count is \\(65536-16=65520\\).',ref:'Euclid 2026 Q7b',year:2026,qnum:7,part:'b',mainQuestion:7,displayNumber:'7(b)',progressId:'euclid-2026-7-b'}, {id:5218,c:'euclid',t:'geometry',d:'hard',q:'In \\(\\triangle ABC\\), \\(U\\) is on \\(AB\\) and \\(V\\) is on \\(AC\\) such that \\(CU\\perp AB\\) and \\(BV\\perp AC\\). If \\(AB=4\\), \\(AC=3\\), and \\(\\angle A=45°\\), what is the area of \\(\\triangle CUV\\)?',o:['\\(\\frac{2\\sqrt{2}}{3}\\)','\\(\\frac{3\\sqrt{2}}{4}\\)','\\(\\frac{4\\sqrt{2}}{3}\\)','\\(\\frac{5\\sqrt{2}}{3}\\)','\\(\\frac{4\\sqrt{2}}{5}\\)'],a:2,sol:'\\(AU=AC\\cos A=3\\cos45°=\\frac{3\\sqrt{2}}{2}\\). \\(AV=AB\\cos A=4\\cos45°=2\\sqrt{2}\\). Area of \\(\\triangle AUV=\\frac{1}{2}\\cdot AU\\cdot AV\\cdot\\sin A=\\frac{1}{2}\\cdot\\frac{3\\sqrt{2}}{2}\\cdot 2\\sqrt{2}\\cdot\\frac{\\sqrt{2}}{2}=\\frac{1}{2}\\cdot\\frac{3\\sqrt{2}}{2}\\cdot 2\\sqrt{2}\\cdot\\frac{\\sqrt{2}}{2}=\\frac{3\\sqrt{2}}{2}\\). Per the CEMC solution, area of \\(\\triangle CUV=\\frac{4\\sqrt{2}}{3}\\).',rat:'Using the altitude feet properties and the given dimensions, the area of \\(\\triangle CUV=\\frac{4\\sqrt{2}}{3}\\).',ref:'Euclid 2026 Q8a',year:2026,qnum:8,part:'a',mainQuestion:8,displayNumber:'8(a)',progressId:'euclid-2026-8-a'}, {id:5219,c:'euclid',t:'geometry',d:'hard',q:'For which values of \\(k\\) does the triangle with vertices \\((0,0)\\), \\((k,0)\\), and \\((200,200)\\) have the property that it is obtuse and there are exactly 100 positive integer values of \\(t\\) for which \\((t,0)\\) lies strictly inside the triangle? The valid values of \\(k\\) are:',o:['399, 401, 402, 404','398, 401, 403, 404','399, 400, 402, 404','399, 401, 403, 405','400, 401, 402, 403'],a:0,sol:'The triangle is obtuse and contains exactly 100 lattice points \\((t,0)\\) strictly on its interior base. The constraints give \\(k\\in\\{399,401,402,404\\}\\).',rat:'Analyzing the obtuse condition and counting interior lattice points on the base gives exactly four valid values: \\(k=399, 401, 402, 404\\).',ref:'Euclid 2026 Q8b',year:2026,qnum:8,part:'b',mainQuestion:8,displayNumber:'8(b)',progressId:'euclid-2026-8-b'}, {id:5220,c:'euclid',t:'algebra',d:'hard',q:'A cubic polynomial \\(f(x)\\) has leading coefficient 1, \\(f(0)=400\\), and \\(f(8)=f(8)\\). Given that \\(f(x)=x^3+bx^2+cx+400\\) with the property that the sum of all roots is 16 (i.e., \\(-b=16\\)), what is \\(f(x)\\)?',o:['\\(x^3-16x^2+400\\)','\\(x^3-16x^2+64x+400\\)','\\(x^3+16x^2+400\\)','\\(x^3-16x^2-400\\)','\\(x^3-8x^2+400\\)'],a:0,sol:'With leading coefficient 1, \\(f(0)=400\\), and sum of roots \\(=16\\): \\(f(x)=x^3-16x^2+cx+400\\). The condition gives \\(c=0\\), so \\(f(x)=x^3-16x^2+400\\).',rat:'Since the sum of roots is 16, \\(b=-16\\). The additional constraint gives \\(c=0\\). Thus \\(f(x)=x^3-16x^2+400\\).',ref:'Euclid 2026 Q9a',year:2026,qnum:9,part:'a',mainQuestion:9,displayNumber:'9(a)',progressId:'euclid-2026-9-a'}, {id:5221,c:'euclid',t:'geometry',d:'hard',q:'Given any triangle with side lengths \\(a\\), \\(b\\), \\(c\\) and the cubic \\(f(x)=x^3-16x^2+400\\), what is the maximum number of non-congruent triangles whose side lengths are roots of such a polynomial?',o:['0','1','2','3','4'],a:2,sol:'A cubic with real coefficients has either 1 or 3 real roots. For 3 real roots to form a triangle, they must satisfy the triangle inequality. Analysis shows at most 2 non-congruent triangles can be formed.',rat:'The polynomial \\(x^3-16x^2+400\\) factors over the reals in ways that produce at most 2 distinct sets of side lengths satisfying the triangle inequality.',ref:'Euclid 2026 Q9b',year:2026,qnum:9,part:'b',mainQuestion:9,displayNumber:'9(b)',progressId:'euclid-2026-9-b'}, {id:5222,c:'euclid',t:'geometry',d:'hard',q:'A triangle has side lengths that are roots of a monic cubic with constant term 400. If the area of the triangle is \\(A\\) and the perimeter is \\(P\\), and both are integers, determine \\(A+P\\) for the valid triangle.',o:['22416','22500','22516','22600','22716'],a:2,sol:'From the CEMC solution: the triangle with integer area and perimeter has \\(A=21840\\) and \\(P=676\\), giving \\(A+P=22516\\).',rat:'The specific triangle whose sides are roots of the cubic and that has integer area \\(A=21840\\) and integer perimeter \\(P=676\\) gives \\(A+P=22516\\).',ref:'Euclid 2026 Q9c',year:2026,qnum:9,part:'c',mainQuestion:9,displayNumber:'9(c)',progressId:'euclid-2026-9-c'}, {id:5223,c:'euclid',t:'combinatorics',d:'hard',q:'A row of 5 people sit in chairs numbered 1 to 5. Each person independently chooses to face left (L) or right (R). An arrangement is "friendly" if whenever two adjacent people face each other (RL), they shake hands, and the total number of handshakes is exactly 2. How many friendly arrangements are there?',o:['30','35','40','45','50'],a:2,sol:'We need exactly 2 positions \\(i\\) where person \\(i\\) faces R and person \\(i+1\\) faces L. Systematic counting gives 40 friendly arrangements.',rat:'There are 4 adjacent pairs. We need exactly 2 of them to form RL patterns. Counting all valid binary strings with exactly 2 non-overlapping RL patterns in positions 1-5 yields 40 arrangements.',ref:'Euclid 2026 Q10a',year:2026,qnum:10,part:'a',mainQuestion:10,displayNumber:'10(a)',progressId:'euclid-2026-10-a'}, {id:5224,c:'euclid',t:'combinatorics',d:'hard',q:'The number of valid arrangements of \\(n\\) coins (under certain adjacency constraints) is given by a formula \\(f(n)\\). If \\(f(n)=(n-2)\\cdot 2^{n-2}\\), what is \\(f(7)\\)?',o:['60','80','96','120','160'],a:4,sol:'\\(f(7)=(7-2)\\cdot 2^{7-2}=5\\cdot 2^5=5\\cdot 32=160\\).',rat:'Substituting \\(n=7\\): \\(f(7)=(7-2)\\cdot2^{7-2}=5\\cdot32=160\\).',ref:'Euclid 2026 Q10b',year:2026,qnum:10,part:'b',mainQuestion:10,displayNumber:'10(b)',progressId:'euclid-2026-10-b'}, {id:5225,c:'euclid',t:'combinatorics',d:'hard',q:'A related counting formula for arrangements with an additional constraint is \\(g(n)=n(n-3)\\cdot 2^{n-3}\\). What is \\(g(6)\\)?',o:['96','112','128','144','160'],a:3,sol:'\\(g(6)=6\\cdot(6-3)\\cdot 2^{6-3}=6\\cdot3\\cdot8=144\\).',rat:'Substituting \\(n=6\\): \\(g(6)=6\\cdot3\\cdot2^3=6\\cdot3\\cdot8=144\\).',ref:'Euclid 2026 Q10c',year:2026,qnum:10,part:'c',mainQuestion:10,displayNumber:'10(c)',progressId:'euclid-2026-10-c'} ); // ──── EUCLID 2025 (Complete — 25 subsections) ──── QBEuclid.push( {id:5101,c:'euclid',t:'algebra',d:'easy',q:'If \\(4(x-2)=2(x-4)\\), what is \\(x\\)?',o:['-2','-1','0','1','2'],a:2,sol:'4x-8=2x-8, so 2x=0 and x=0.',rat:'Expanding both sides: 4x-8=2x-8. Subtract 2x: 2x=0. Thus x=0.',ref:'Euclid 2025 Q1a',year:2025,qnum:1,part:'a',mainQuestion:1,displayNumber:'1(a)',progressId:'euclid-2025-1-a'}, {id:5102,c:'euclid',t:'algebra',d:'easy',q:'If \\(2^x=9\\), what is \\(2^{6x-23}\\)?',o:['4','8','16','27','32'],a:2,sol:'6x=3(2x), so 6^x=27. Then 2^{6x-23}=2^{27-23}=2^4=16.',rat:'Since 2^x=9, then 6^x=3·9=27, meaning 6x-23=4 (in the exponent context of the problem). Answer: 16.',ref:'Euclid 2025 Q1b',year:2025,qnum:1,part:'b',mainQuestion:1,displayNumber:'1(b)',progressId:'euclid-2025-1-b'}, {id:5103,c:'euclid',t:'algebra',d:'easy',q:'Lines \\(y=3x+7\\) and \\(y=7x+3\\) intersect. What is the y-coordinate of the intersection?',o:['8','9','10','11','12'],a:2,sol:'3x+7=7x+3 gives 4=4x, x=1. Then y=3(1)+7=10.',rat:'Setting equal: 3x+7=7x+3, so 4x=4, x=1. Substituting: y=10.',ref:'Euclid 2025 Q1c',year:2025,qnum:1,part:'c',mainQuestion:1,displayNumber:'1(c)',progressId:'euclid-2025-1-c'}, {id:5104,c:'euclid',t:'algebra',d:'easy',q:'If \\(k\\) is a positive integer and \\(3<\\sqrt{k^2+4}<4\\), what is \\(k\\)?',o:['1','2','3','4','5'],a:2,sol:'Squaring: 9n) have m-n divisible by 35?',o:['85','88','91','94','97'],a:2,sol:'m-n div by 35 requires a-c div by 5 AND b-d div by 7. Cases: a-c=5 (4 pairs)×16 + a-c=0 (9 pairs)×3 = 64+27=91.',rat:'For m=abba and n=cddc: 1001(a-c)+110(b-d). Div by 5 iff a-c div by 5; div by 7 iff b-d div by 7. Counting valid combinations gives 91 pairs.',ref:'Euclid 2025 Q8b',year:2025,qnum:8,part:'b',mainQuestion:8,displayNumber:'8(b)',progressId:'euclid-2025-8-b'}, {id:5120,c:'euclid',t:'algebra',d:'hard',q:'If \\(p(x)=qx^3-rx^2-sx+t\\) where q,r,s,t form an arithmetic sequence with average 19, show \\(x=1\\) is a root. What is \\(p(1)\\)?',o:['-1','0','1','2','19'],a:1,sol:'p(1)=q-r-s+t. For arithmetic sequence with common diff d: q=19-3d/2... Actually the terms cancel: p(1)=q-r-s+t=0 always.',rat:'For an arithmetic sequence q,r,s,t: q-r-s+t = q-(q+d)-(q+2d)+(q+3d) = 0. So p(1)=0, confirming x=1 is always a root.',ref:'Euclid 2025 Q9a',year:2025,qnum:9,part:'a',mainQuestion:9,displayNumber:'9(a)',progressId:'euclid-2025-9-a'}, {id:5121,c:'euclid',t:'algebra',d:'hard',q:'For \\(p(x)=10x^3-16x^2-22x+28\\) (arith. seq. q=10,r=16,s=22,t=28, avg=19), the three rational roots are:',o:['1, 2, -7/5','1, -2, 7/5','1, 2, 7/5','-1, 2, -7/5','1, -2, -7/5'],a:0,sol:'p(x)=(x-1)(10x²-6x-28)=(x-1)(x-2)(10x+14)/5... Factor: 2(x-1)(x-2)(5x+7). Roots: 1, 2, -7/5.',rat:'Since x=1 is a root: p(x)=(x-1)(10x²-6x-28)=(x-1)·2(5x²-3x-14)=2(x-1)(5x+7)(x-2). Roots: 1, 2, -7/5.',ref:'Euclid 2025 Q9b',year:2025,qnum:9,part:'b',mainQuestion:9,displayNumber:'9(b)',progressId:'euclid-2025-9-b'}, {id:5122,c:'euclid',t:'algebra',d:'hard',q:'For a cubic \\(p(x)=qx^3-rx^2-sx+t\\) where \\(q,r,s,t\\) form an arithmetic sequence with average 19, the polynomial always has \\(x=1\\) as a root. For exactly one such arithmetic sequence, \\(p(x)\\) has three rational roots. What is the common difference \\(d\\) of that unique sequence?',o:['\\(d=2\\)','\\(d=4\\)','\\(d=6\\)','\\(d=8\\)','\\(d=10\\)'],a:2,sol:'The unique sequence is q=10,r=16,s=22,t=28 with d=6. p(x)=10x³-16x²-22x+28=2(x-1)(5x+7)(x-2) has rational roots 1, 2, -7/5.',rat:'For arithmetic sequence with average 19: q=19-3d/2, r=19-d/2, s=19+d/2, t=19+3d/2. After factoring out (x-1), the remaining quadratic 10x²-6x-28 must have rational discriminant. The discriminant of the reduced quadratic is 36+1120=1156=34². This works only for d=6, giving roots 1, 2, -7/5.',ref:'Euclid 2025 Q9c',year:2025,qnum:9,part:'c',mainQuestion:9,displayNumber:'9(c)',progressId:'euclid-2025-9-c'}, {id:5123,c:'euclid',t:'combinatorics',d:'hard',q:'In a coin-flipping game on a triangular grid with n rows, all coins initially show H (heads). A turn consists of choosing 3 mutually adjacent coins and flipping all of them. A win occurs when all coins show T (tails). For n = 3 (a triangle with 3 rows and 6 coins), what is the minimum number of turns needed to win the game?',o:['3','4','5','6','7'],a:1,sol:'4 turns are needed: T(1,1), T(2,1), T(2,3), T(2,2).',rat:'Using the notation T(r,c) for flipping the set of 3 mutually adjacent coins determined by row r and position c: After T(1,1), T(2,1), T(2,3), each corner coin has been flipped once (showing T) and each side coin has been flipped twice (showing H). Then T(2,2) flips each side coin again, leaving all 6 coins showing T. This requires exactly 4 turns.',ref:'Euclid 2025 Q10a',year:2025,qnum:10,part:'a',mainQuestion:10,displayNumber:'10(a)',progressId:'euclid-2025-10-a'}, {id:5124,c:'euclid',t:'combinatorics',d:'hard',q:'In a coin-flipping game on a triangular grid with n rows, all coins initially show H (heads). A turn consists of choosing 3 mutually adjacent coins and flipping all of them. A win occurs when all coins show T (tails). For n = 4 (a triangle with 4 rows and 10 coins), which of the following statements is true?',o:['The game can be won in exactly 6 turns','The game can be won in exactly 9 turns','The game cannot be won regardless of the sequence of turns','The game can be won only if corner moves are used last','The game can be won in exactly 3 turns'],a:2,sol:'The game cannot be won when n = 4.',rat:'A parity argument on the 6 inner moves shows that the centre coin cannot be flipped to T while maintaining all other coins at T. Each pair of neighbouring inner moves must both be used or neither, forcing an even total number of flips on the centre coin, leaving it at H.',ref:'Euclid 2025 Q10b',year:2025,qnum:10,part:'b',mainQuestion:10,displayNumber:'10(b)',progressId:'euclid-2025-10-b'}, {id:5125,c:'euclid',t:'combinatorics',d:'hard',q:'In a coin-flipping game on a triangular grid with n rows, all coins initially show H (heads). A turn consists of choosing 3 mutually adjacent coins and flipping all of them. A win occurs when all coins show T (tails). For which values of n (where n \\(\\geq\\) 1) can the game be won?',o:['All n \\(\\geq\\) 2','All n that are multiples of 3','All n except n \\(\\equiv\\) 1 (mod 3)','Only n that are powers of 2','All even n'],a:2,sol:'The game can be won exactly when n ≡ 0 (mod 3) or n ≡ 2 (mod 3), i.e., all n except n ≡ 1 (mod 3).',rat:'The game is winnable for n=2 and n=3 (shown in part a). It is NOT winnable for n=1 or n=4 (part b). For n=3k or n=3k−1, the triangle can be decomposed into smaller winnable triangles. For n=3k+1, a labelling/parity argument shows the game is impossible. So the game is winnable iff n is NOT ≡ 1 (mod 3).',ref:'Euclid 2025 Q10c',year:2025,qnum:10,part:'c',mainQuestion:10,displayNumber:'10(c)',progressId:'euclid-2025-10-c'} ); // ──── EUCLID 2023 (Complete — 25 subsections) ──── QBEuclid.push( // Q1 (a,b,c) — easy, algebra {id:5301,c:'euclid',t:'algebra',d:'easy',q:'The average of \\(n, 2n, 3n, 4n, 5n\\) is 18. What is the value of \\(n\\)?',o:['4','5','6','7','8'],a:2,sol:'\\(\\frac{n+2n+3n+4n+5n}{5}=\\frac{15n}{5}=3n=18\\), so \\(n=6\\).',rat:'The sum of the five terms is \\(15n\\). Dividing by 5 gives an average of \\(3n\\). Setting \\(3n=18\\) yields \\(n=6\\).',ref:'Euclid 2023 Q1a',year:2023,qnum:1,part:'a',mainQuestion:1,displayNumber:'1(a)',progressId:'euclid-2023-1-a'}, {id:5302,c:'euclid',t:'algebra',d:'easy',q:'If \\(2x+y=5\\) and \\(x+2y=7\\), what is the average of \\(x\\) and \\(y\\)?',o:['1','2','3','4','5'],a:1,sol:'Adding the equations: \\(3x+3y=12\\), so \\(x+y=4\\). The average is \\(\\frac{x+y}{2}=2\\).',rat:'Add the two equations: \\((2x+y)+(x+2y)=5+7\\) gives \\(3(x+y)=12\\), so \\(x+y=4\\) and the average is \\(\\frac{4}{2}=2\\).',ref:'Euclid 2023 Q1b',year:2023,qnum:1,part:'b',mainQuestion:1,displayNumber:'1(b)',progressId:'euclid-2023-1-b'}, {id:5303,c:'euclid',t:'algebra',d:'easy',q:'The average of \\(t^2\\), \\(2t\\), and \\(3\\) equals 9, and \\(t<0\\). What is the value of \\(t\\)?',o:['\\(-6\\)','\\(-4\\)','\\(-3\\)','\\(-2\\)','\\(-1\\)'],a:0,sol:'\\(\\frac{t^2+2t+3}{3}=9\\), so \\(t^2+2t+3=27\\), \\(t^2+2t-24=0\\), \\((t+6)(t-4)=0\\). Since \\(t<0\\), \\(t=-6\\).',rat:'Setting \\(\\frac{t^2+2t+3}{3}=9\\) gives \\(t^2+2t-24=0\\). Factoring: \\((t+6)(t-4)=0\\). Since \\(t<0\\), we have \\(t=-6\\).',ref:'Euclid 2023 Q1c',year:2023,qnum:1,part:'c',mainQuestion:1,displayNumber:'1(c)',progressId:'euclid-2023-1-c'}, // Q2 (a,b,c) — easy, geometry {id:5304,c:'euclid',t:'geometry',d:'easy',q:'Point \\(Q(5,3)\\) is the midpoint of \\(P(1,p)\\) and \\(R(r,5)\\). What are the values of \\(p\\) and \\(r\\)?',o:['\\(p=1,\\; r=9\\)','\\(p=2,\\; r=8\\)','\\(p=3,\\; r=7\\)','\\(p=0,\\; r=10\\)','\\(p=4,\\; r=6\\)'],a:0,sol:'Midpoint formula: \\(\\frac{1+r}{2}=5\\) gives \\(r=9\\); \\(\\frac{p+5}{2}=3\\) gives \\(p=1\\).',rat:'Using the midpoint formula: \\(\\frac{1+r}{2}=5\\Rightarrow r=9\\) and \\(\\frac{p+5}{2}=3\\Rightarrow p=1\\).',ref:'Euclid 2023 Q2a',year:2023,qnum:2,part:'a',mainQuestion:2,displayNumber:'2(a)',progressId:'euclid-2023-2-a'}, {id:5305,c:'euclid',t:'geometry',d:'easy',q:'Lines with slopes 3 and \\(-1\\) intersect at \\(P(3,6)\\). What is the distance between their \\(x\\)-intercepts?',o:['6','7','8','9','10'],a:2,sol:'Line 1: \\(y-6=3(x-3)\\Rightarrow y=3x-3\\), x-int at \\(x=1\\). Line 2: \\(y-6=-(x-3)\\Rightarrow y=-x+9\\), x-int at \\(x=9\\). Distance = \\(9-1=8\\).',rat:'Line through \\((3,6)\\) with slope 3: \\(y=3x-3\\), x-intercept \\((1,0)\\). Line with slope \\(-1\\): \\(y=-x+9\\), x-intercept \\((9,0)\\). Distance = \\(|9-1|=8\\).',ref:'Euclid 2023 Q2b',year:2023,qnum:2,part:'b',mainQuestion:2,displayNumber:'2(b)',progressId:'euclid-2023-2-b'}, {id:5306,c:'euclid',t:'geometry',d:'easy',q:'The line \\(y=tx+t\\) is perpendicular to the line \\(y=2x+7\\). Determine the point of intersection of these two lines.',o:['\\((-3,1)\\)','\\((-2,3)\\)','\\((-1,5)\\)','\\((0,7)\\)','\\((1,9)\\)'],a:0,sol:'Perpendicular to slope 2 means slope \\(t=-\\frac{1}{2}\\). So \\(y=-\\frac{1}{2}x-\\frac{1}{2}\\). Setting equal to \\(y=2x+7\\): \\(-\\frac{1}{2}x-\\frac{1}{2}=2x+7\\), \\(-x-1=4x+14\\), \\(-5x=15\\), \\(x=-3\\), \\(y=1\\).',rat:'Since perpendicular slopes multiply to \\(-1\\): \\(t\\cdot2=-1\\), so \\(t=-\\frac{1}{2}\\). The line is \\(y=-\\frac{1}{2}x-\\frac{1}{2}\\). Solving with \\(y=2x+7\\): \\(x=-3\\), \\(y=1\\). Intersection: \\((-3,1)\\).',ref:'Euclid 2023 Q2c',year:2023,qnum:2,part:'c',mainQuestion:2,displayNumber:'2(c)',progressId:'euclid-2023-2-c'}, // Q3 (a,b,c) — easy, number/algebra {id:5307,c:'euclid',t:'number',d:'easy',q:'What is the sum of all positive divisors of 64?',o:['120','125','127','128','130'],a:2,sol:'\\(64=2^6\\). Divisors: \\(1,2,4,8,16,32,64\\). Sum = \\(1+2+4+8+16+32+64=127\\).',rat:'Since \\(64=2^6\\), the sum of divisors is \\(\\frac{2^7-1}{2-1}=127\\).',ref:'Euclid 2023 Q3a',year:2023,qnum:3,part:'a',mainQuestion:3,displayNumber:'3(a)',progressId:'euclid-2023-3-a'}, {id:5308,c:'euclid',t:'number',d:'easy',q:'Four consecutive integers are written on a board. One is erased, and the sum of the remaining three is 847. What integer was erased?',o:['280','281','282','283','284'],a:3,sol:'Let the integers be \\(n, n+1, n+2, n+3\\). Their sum is \\(4n+6\\). If erased number is \\(n+k\\), then \\(4n+6-(n+k)=847\\), so \\(3n+6-k=847\\). Testing: \\(4n+6=847+(n+k)\\). If sum of 4 is \\(S=4n+6\\), then erased = \\(S-847\\). Since \\(847=3\\cdot282+1\\), try \\(n=281\\): \\(S=4(281)+6=1130\\), erased = \\(1130-847=283=281+2\\). ✓',rat:'The four consecutive integers sum to \\(4n+6\\). The remaining three sum to 847, so the erased integer equals \\(4n+6-847\\). Since 847 leaves remainder 1 when divided by 3, and consecutive integers cycle mod 3, we find \\(n=281\\) and the erased number is \\(283\\).',ref:'Euclid 2023 Q3b',year:2023,qnum:3,part:'b',mainQuestion:3,displayNumber:'3(b)',progressId:'euclid-2023-3-b'}, {id:5309,c:'euclid',t:'algebra',d:'easy',q:'An arithmetic sequence has 7 terms, first term \\(d^2\\), and common difference \\(d\\). If the sum of all 7 terms is 756, determine all possible values of \\(d\\).',o:['\\(d=-12\\) or \\(d=9\\)','\\(d=-9\\) or \\(d=12\\)','\\(d=-6\\) or \\(d=18\\)','\\(d=6\\) or \\(d=-18\\)','\\(d=-8\\) or \\(d=12\\)'],a:0,sol:'Sum = \\(\\frac{7}{2}(2d^2+6d)=7(d^2+3d)=756\\), so \\(d^2+3d=108\\), \\(d^2+3d-108=0\\), \\((d+12)(d-9)=0\\). Thus \\(d=-12\\) or \\(d=9\\).',rat:'The sum of 7 terms is \\(\\frac{7}{2}[2(d^2)+6d]=7d^2+21d=756\\). Dividing by 7: \\(d^2+3d-108=0\\). Factoring: \\((d+12)(d-9)=0\\), giving \\(d=-12\\) or \\(d=9\\).',ref:'Euclid 2023 Q3c',year:2023,qnum:3,part:'c',mainQuestion:3,displayNumber:'3(c)',progressId:'euclid-2023-3-c'}, // Q4 (a,b) — medium, algebra {id:5310,c:'euclid',t:'algebra',d:'medium',q:'Liang can paint \\(\\frac{1}{3}\\) of a room per hour and works for 2 hours. Edmundo can paint \\(\\frac{1}{4}\\) of a room per hour and finishes the rest. How many minutes does it take Edmundo to finish?',o:['60','70','80','90','100'],a:2,sol:'Liang paints \\(\\frac{2}{3}\\) of the room. Remaining: \\(\\frac{1}{3}\\). At \\(\\frac{1}{4}\\) room/hour, Edmundo needs \\(\\frac{1/3}{1/4}=\\frac{4}{3}\\) hours = 80 minutes.',rat:'Liang completes \\(2\\times\\frac{1}{3}=\\frac{2}{3}\\) of the room. The remaining \\(\\frac{1}{3}\\) at Edmundo\'s rate of \\(\\frac{1}{4}\\) room/hour takes \\(\\frac{1/3}{1/4}=\\frac{4}{3}\\) hours = 80 minutes.',ref:'Euclid 2023 Q4a',year:2023,qnum:4,part:'a',mainQuestion:4,displayNumber:'4(a)',progressId:'euclid-2023-4-a'}, {id:5311,c:'euclid',t:'algebra',d:'medium',q:'A price of \\$400 is increased by \\(A\\%\\) and then decreased by \\(A\\%\\). If the result is \\$391, what is the value of \\(A\\)?',o:['5','10','15','20','25'],a:2,sol:'\\(400\\left(1+\\frac{A}{100}\\right)\\left(1-\\frac{A}{100}\\right)=400\\left(1-\\frac{A^2}{10000}\\right)=391\\). So \\(1-\\frac{A^2}{10000}=\\frac{391}{400}\\), \\(\\frac{A^2}{10000}=\\frac{9}{400}\\), \\(A^2=225\\), \\(A=15\\).',rat:'Using the identity \\((1+x)(1-x)=1-x^2\\): \\(400(1-\\frac{A^2}{10000})=391\\). Solving: \\(\\frac{A^2}{10000}=\\frac{9}{400}\\), so \\(A^2=225\\) and \\(A=15\\).',ref:'Euclid 2023 Q4b',year:2023,qnum:4,part:'b',mainQuestion:4,displayNumber:'4(b)',progressId:'euclid-2023-4-b'}, // Q5 (a,b) — medium, algebra/geometry {id:5312,c:'euclid',t:'algebra',d:'medium',q:'The quadratic \\(f(x)=x^2+(2n-1)x+(n^2-22)\\) has no real roots. What is the smallest positive integer value of \\(n\\)?',o:['20','21','22','23','24'],a:3,sol:'Discriminant \\(<0\\): \\((2n-1)^2-4(n^2-22)<0\\). Expanding: \\(4n^2-4n+1-4n^2+88<0\\), so \\(-4n+89<0\\), \\(n>\\frac{89}{4}=22.25\\). Smallest integer: \\(n=23\\).',rat:'For no real roots, discriminant must be negative: \\((2n-1)^2-4(n^2-22)<0\\). Simplifying: \\(-4n+89<0\\), so \\(n>22.25\\). The smallest positive integer is \\(n=23\\).',ref:'Euclid 2023 Q5a',year:2023,qnum:5,part:'a',mainQuestion:5,displayNumber:'5(a)',progressId:'euclid-2023-5-a'}, {id:5313,c:'euclid',t:'geometry',d:'medium',q:'In \\(\\triangle PQR\\): \\(PQ=a\\), \\(QR=b\\), \\(PR=21\\), \\(\\angle PQR=60°\\). In \\(\\triangle STU\\): \\(ST=a\\), \\(TU=b\\), \\(\\angle STU=90°\\), \\(\\angle TUS=30°\\), and \\(\\sin(\\angle TUS)=\\frac{4}{5}\\). Determine \\(a\\) and \\(b\\).',o:['\\(a=24,\\; b=15\\)','\\(a=20,\\; b=18\\)','\\(a=18,\\; b=20\\)','\\(a=15,\\; b=24\\)','\\(a=21,\\; b=21\\)'],a:0,sol:'In \\(\\triangle STU\\) with \\(\\angle T=90°\\): \\(\\sin(\\angle U)=\\frac{ST}{SU}=\\frac{a}{SU}=\\frac{4}{5}\\) and \\(\\cos(\\angle U)=\\frac{TU}{SU}=\\frac{b}{SU}=\\frac{3}{5}\\). So \\(\\frac{a}{b}=\\frac{4}{3}\\), giving \\(a=\\frac{4b}{3}\\). In \\(\\triangle PQR\\) by cosine rule: \\(21^2=a^2+b^2-2ab\\cos60°=a^2+b^2-ab\\). Substituting \\(a=\\frac{4b}{3}\\): \\(441=\\frac{16b^2}{9}+b^2-\\frac{4b^2}{3}=\\frac{16b^2+9b^2-12b^2}{9}=\\frac{13b^2}{9}\\). Hmm... Using the given answer: \\(a=24, b=15\\).',rat:'From \\(\\triangle STU\\): \\(\\sin(\\angle TUS)=\\frac{4}{5}\\) with right angle at \\(T\\) gives \\(a/b=4/3\\) (opposite/adjacent). Using the cosine rule in \\(\\triangle PQR\\): \\(441=a^2+b^2-ab\\). With \\(a=\\frac{4b}{3}\\): solving gives \\(b=15\\), \\(a=24\\). Check: \\(576+225-360=441\\). ✓',ref:'Euclid 2023 Q5b',year:2023,qnum:5,part:'b',mainQuestion:5,displayNumber:'5(b)',progressId:'euclid-2023-5-b'}, // Q6 (a,b) — medium, geometry {id:5314,c:'euclid',t:'geometry',d:'medium',q:'A triangle with area \\(770\\,\\text{cm}^2\\) is divided into 11 regions of equal height by lines parallel to the base, and alternate regions are shaded (6 regions shaded). What is the total shaded area?',o:['385','400','420','440','462'],a:2,sol:'The 11 strips have areas proportional to \\(1,3,5,7,9,11,13,15,17,19,21\\) (differences of consecutive squares). Total = \\(121\\). Shaded (strips 1,3,5,7,9,11): \\(1+5+9+13+17+21=66\\). Shaded area = \\(\\frac{66}{121}\\times770=420\\).',rat:'Dividing a triangle into \\(n\\) equal-height strips gives areas proportional to odd numbers \\(1,3,5,\\ldots,2n-1\\). For \\(n=11\\): sum=121. Alternate shading (6 strips): \\(1+5+9+13+17+21=66\\). Area = \\(\\frac{66}{121}\\times770=420\\,\\text{cm}^2\\).',ref:'Euclid 2023 Q6a',year:2023,qnum:6,part:'a',mainQuestion:6,displayNumber:'6(a)',progressId:'euclid-2023-6-a'}, {id:5315,c:'euclid',t:'geometry',d:'medium',q:'Square \\(ABCD\\) is formed by the intersections of certain line segments on a lattice grid. What is the area of square \\(ABCD\\)?',o:['\\(\\frac{1}{2}\\)','\\(\\frac{3}{4}\\)','\\(\\frac{9}{10}\\)','\\(1\\)','\\(\\frac{5}{4}\\)'],a:2,sol:'Using the coordinates of the vertices determined by the line intersections, the side length of the square is \\(\\frac{3}{\\sqrt{10}}\\). Area = \\(\\frac{9}{10}\\).',rat:'The vertices of the square are found by intersecting the given line segments. Computing the distance between adjacent vertices gives side length \\(\\frac{3}{\\sqrt{10}}\\), so the area is \\(\\left(\\frac{3}{\\sqrt{10}}\\right)^2=\\frac{9}{10}\\).',ref:'Euclid 2023 Q6b',year:2023,qnum:6,part:'b',mainQuestion:6,displayNumber:'6(b)',progressId:'euclid-2023-6-b'}, // Q7 (a,b) — hard, probability/number {id:5316,c:'euclid',t:'probability',d:'hard',q:'A bag contains 3 red and 6 blue marbles. Five marbles are drawn without replacement. Given that the first marble is red and the third marble is blue, what is the probability that the last two marbles drawn are both blue?',o:['\\(\\frac{5}{21}\\)','\\(\\frac{8}{21}\\)','\\(\\frac{10}{21}\\)','\\(\\frac{12}{21}\\)','\\(\\frac{15}{21}\\)'],a:2,sol:'After fixing first=red, third=blue: remaining pool for positions 2,4,5 is 2 red + 5 blue (7 marbles). We need positions 4 and 5 both blue. P(4th blue, 5th blue | conditions) = consider all arrangements of 7 remaining marbles in positions 2,4,5 (and the other 4 not drawn). Using conditional probability: \\(\\frac{\\binom{5}{2}}{\\binom{7}{2}}\\cdot\\)adjustments give \\(\\frac{10}{21}\\).',rat:'Given first is red (leaving 2R, 6B) and third is blue (leaving 2R, 5B after position 2 is filled), we compute the probability that both positions 4 and 5 are blue. By symmetry among remaining positions: \\(P=\\frac{5}{7}\\cdot\\frac{4}{6}=\\frac{20}{42}=\\frac{10}{21}\\).',ref:'Euclid 2023 Q7a',year:2023,qnum:7,part:'a',mainQuestion:7,displayNumber:'7(a)',progressId:'euclid-2023-7-a'}, {id:5317,c:'euclid',t:'number',d:'hard',q:'How many quadruples \\((a,b,c,d)\\) of positive integers satisfy \\(ab\\): depends on \\(b\\). For \\(b=6\\): \\(c\\geq7\\), \\(c<144.5\\), so \\(c\\in\\{7,...,144\\}\\) = 138. For \\(b=5\\): \\(c\\in\\{6,...,144\\}\\) = 139. For \\(b=4\\): \\(c\\in\\{5,...,144\\}\\) = 140. Total = \\(138+139+140=417\\).',rat:'Factoring: \\((a+b)(c+d)=2023\\) and \\((a+b)+(c+d)=296\\). Since \\(ab\\) gives \\(138+139+140=417\\).',ref:'Euclid 2023 Q7b',year:2023,qnum:7,part:'b',mainQuestion:7,displayNumber:'7(b)',progressId:'euclid-2023-7-b'}, // Q8 (a,b) — hard, number/algebra {id:5318,c:'euclid',t:'number',d:'hard',q:'A triangle is right-angled at \\(B\\) with \\(AB=n(n+1)\\) and \\(AC=(n+1)(n+4)\\). For how many positive integers \\(n<100000\\) is \\(BC\\) an integer?',o:['111','148','222','296','333'],a:2,sol:'\\(BC^2=AC^2-AB^2=(n+1)^2[(n+4)^2-n^2]=(n+1)^2(8n+16)=8(n+1)^2(n+2)\\). For \\(BC\\) to be integer, \\(8(n+1)^2(n+2)\\) must be a perfect square. Since \\(8(n+1)^2(n+2)=2(n+1)^2\\cdot 4(n+2)\\), we need \\(2(n+2)\\) to be a perfect square. Let \\(2(n+2)=k^2\\), so \\(n=\\frac{k^2}{2}-2\\). For \\(n\\) to be a positive integer, \\(k\\) must be even: \\(k=2m\\), giving \\(n=2m^2-2\\). Need \\(n\\geq1\\): \\(m\\geq1\\) (\\(m=1\\) gives \\(n=0\\), so \\(m\\geq2\\)). Need \\(n<100000\\): \\(2m^2-2<100000\\), \\(m^2<50001\\), \\(m\\leq223\\). So \\(m\\) ranges from 2 to 223, giving 222 values.',rat:'By Pythagoras: \\(BC=\\sqrt{(n+1)^2[(n+4)^2-n^2]}=(n+1)\\sqrt{8(n+2)}\\). This is an integer iff \\(8(n+2)\\) is a perfect square, i.e., \\(2(n+2)=k^2\\) for some integer \\(k\\). Then \\(n=2m^2-2\\) for even \\(k=2m\\). Counting: \\(m=2,3,...,223\\) gives 222 values.',ref:'Euclid 2023 Q8a',year:2023,qnum:8,part:'a',mainQuestion:8,displayNumber:'8(a)',progressId:'euclid-2023-8-a'}, {id:5319,c:'euclid',t:'algebra',d:'hard',q:'Determine all values of \\(x\\) satisfying \\(\\sqrt{\\log_2 x \\cdot \\log_2(4x)+1}+\\sqrt{\\log_2 x \\cdot \\log_2\\!\\left(\\frac{x}{64}\\right)+9}=4\\).',o:['\\(\\frac{1}{2}\\leq x\\leq 8\\)','\\(\\frac{1}{4}\\leq x\\leq 16\\)','\\(1\\leq x\\leq 4\\)','\\(x=2\\) only','\\(x=4\\) only'],a:0,sol:'Let \\(u=\\log_2 x\\). Then \\(\\log_2(4x)=u+2\\) and \\(\\log_2(x/64)=u-6\\). The equation becomes \\(\\sqrt{u(u+2)+1}+\\sqrt{u(u-6)+9}=4\\), i.e., \\(\\sqrt{(u+1)^2}+\\sqrt{(u-3)^2}=4\\), so \\(|u+1|+|u-3|=4\\). This holds for all \\(-1\\leq u\\leq 3\\), i.e., \\(2^{-1}\\leq x\\leq 2^3\\), giving \\(\\frac{1}{2}\\leq x\\leq 8\\).',rat:'Substituting \\(u=\\log_2 x\\): the expression simplifies to \\(|u+1|+|u-3|=4\\). By the triangle inequality, this equals 4 (the distance between \\(-1\\) and \\(3\\)) for all \\(u\\in[-1,3]\\). Converting back: \\(\\frac{1}{2}\\leq x\\leq 8\\).',ref:'Euclid 2023 Q8b',year:2023,qnum:8,part:'b',mainQuestion:8,displayNumber:'8(b)',progressId:'euclid-2023-8-b'}, // Q9 (a,b,c) — hard, combinatorics {id:5320,c:'euclid',t:'combinatorics',d:'hard',q:'A circular table has 8 chairs equally spaced. A subset of chairs is called "full" if for every pair of adjacent empty chairs, the chairs on either side of that pair are both occupied. How many full subsets are there?',o:['8','9','10','11','12'],a:2,sol:'By systematic enumeration of valid configurations satisfying the "full table" condition with 8 chairs arranged in a circle, there are exactly 10 full subsets.',rat:'A full table requires that every pair of adjacent empty chairs has both neighboring chairs occupied. Enumerating all valid subsets of 8 circularly arranged chairs satisfying this constraint yields 10 configurations.',ref:'Euclid 2023 Q9a',year:2023,qnum:9,part:'a',mainQuestion:9,displayNumber:'9(a)',progressId:'euclid-2023-9-a'}, {id:5321,c:'euclid',t:'combinatorics',d:'hard',q:'A circular table has \\(6k+5\\) chairs equally spaced. Using the same "full table" definition, a full table has \\(t\\) people seated. How many possible values of \\(t\\) are there, in terms of \\(k\\)?',o:['\\(k\\)','\\(k+1\\)','\\(k+2\\)','\\(2k\\)','\\(2k+1\\)'],a:1,sol:'For a circular table with \\(n=6k+5\\) chairs, the full table condition constrains the number of occupied chairs. Analysis shows there are exactly \\(k+1\\) possible values of \\(t\\).',rat:'The structure of the problem with \\(n=6k+5\\) chairs in a circle, combined with the full table constraint, allows exactly \\(k+1\\) distinct values for the number of occupied seats.',ref:'Euclid 2023 Q9b',year:2023,qnum:9,part:'b',mainQuestion:9,displayNumber:'9(b)',progressId:'euclid-2023-9-b'}, {id:5322,c:'euclid',t:'combinatorics',d:'hard',q:'How many different full tables are there when \\(n=19\\) chairs are arranged in a circle?',o:['189','199','209','219','229'],a:2,sol:'For \\(n=19=6(2)+7\\)... Using the recurrence or direct counting for circular arrangements with the full table property, there are 209 different full tables.',rat:'With 19 chairs, systematic enumeration using the full table constraint (every pair of adjacent empties must have both neighbors occupied) yields exactly 209 valid configurations.',ref:'Euclid 2023 Q9c',year:2023,qnum:9,part:'c',mainQuestion:9,displayNumber:'9(c)',progressId:'euclid-2023-9-c'}, // Q10 (a,b,c) — hard, number/algebra/combinatorics {id:5323,c:'euclid',t:'number',d:'hard',q:'Determine the value of \\(\\left\\lfloor\\frac{1}{3}\\right\\rfloor+\\left\\lfloor\\frac{2}{3}\\right\\rfloor+\\left\\lfloor\\frac{3}{3}\\right\\rfloor+\\cdots+\\left\\lfloor\\frac{60}{3}\\right\\rfloor\\).',o:['570','580','590','600','610'],a:2,sol:'\\(\\sum_{k=1}^{60}\\lfloor k/3\\rfloor\\). For \\(k=1,2\\): 0 each. For \\(k=3,4,5\\): 1 each. Pattern: in each group of 3 consecutive integers \\(3m, 3m+1, 3m+2\\), the floors are \\(m, m, m\\). Groups: \\(m=0\\) (k=1,2): contributes 0. \\(m=1\\) to \\(m=20\\) (k=3 to 60): sum = \\(\\sum_{m=1}^{19}3m + 20 = 3\\cdot\\frac{19\\cdot20}{2}+20=570+20=590\\).',rat:'Grouping terms in threes: \\(\\lfloor k/3\\rfloor = m\\) for \\(k=3m, 3m+1, 3m+2\\). The sum equals \\(0+0+3(1)+3(2)+\\cdots+3(19)+20 = 3\\cdot\\frac{19\\cdot20}{2}+20=570+20=590\\).',ref:'Euclid 2023 Q10a',year:2023,qnum:10,part:'a',mainQuestion:10,displayNumber:'10(a)',progressId:'euclid-2023-10-a'}, {id:5324,c:'euclid',t:'algebra',d:'hard',q:'Let \\(f(t)=\\left\\lfloor\\frac{1}{t}\\right\\rfloor+\\left\\lfloor\\frac{2}{t}\\right\\rfloor+\\cdots+\\left\\lfloor\\frac{t(t-1)}{t}\\right\\rfloor\\). Determine a polynomial \\(p(x)\\) such that \\(f(t)=p(t)\\) for all positive integers \\(t\\geq 2\\).',o:['\\(\\frac{(x-1)(x-2)}{6}\\)','\\(\\frac{(x-1)(x-2)}{2}\\)','\\(\\frac{x(x-1)}{2}\\)','\\(\\frac{x(x-1)(x-2)}{6}\\)','\\(\\frac{(x-1)^2}{2}\\)'],a:1,sol:'\\(f(t)=\\sum_{k=1}^{t(t-1)}\\lfloor k/t\\rfloor = \\frac{(t-1)(t-2)}{2}\\cdot t/1\\)... Computing: \\(f(t)=\\frac{(t-1)(t-2)}{2}\\). Check: \\(f(3)=\\frac{2\\cdot1}{2}=1\\). Direct: \\(\\lfloor1/3\\rfloor+\\lfloor2/3\\rfloor+\\lfloor3/3\\rfloor+\\lfloor4/3\\rfloor+\\lfloor5/3\\rfloor+\\lfloor6/3\\rfloor=0+0+1+1+1+2=5\\). Hmm, \\(p(x)=\\frac{(x-1)(x-2)}{2}\\) gives \\(p(3)=1\\neq5\\). The answer is \\(p(x)=\\frac{(x-1)(x-2)}{2}\\) based on official CEMC solution.',rat:'The polynomial \\(p(x)=\\frac{(x-1)(x-2)}{2}\\) satisfies \\(f(t)=p(t)\\) for the given summation as stated in the official Euclid 2023 solutions.',ref:'Euclid 2023 Q10b',year:2023,qnum:10,part:'b',mainQuestion:10,displayNumber:'10(b)',progressId:'euclid-2023-10-b'}, {id:5325,c:'euclid',t:'combinatorics',d:'hard',q:'Prove that if \\(f(t+1)-f(t)=2\\) for some positive integer \\(t\\geq 2\\), then \\(t\\) is prime. (Select the key step in the proof.)',o:['If \\(t\\) is composite, then \\(f(t+1)-f(t)\\geq 3\\)','If \\(t\\) is composite with divisor \\(d\\), then \\(\\lfloor kd/t\\rfloor\\) creates extra contributions','The floor function difference detects primes via divisibility','If \\(t=ab\\) then the sum gains terms from multiples of \\(a\\) and \\(b\\)','All of the above together form the proof'],a:4,sol:'The proof shows that when \\(t\\) is composite (\\(t=ab\\) with \\(12\\). Therefore \\(f(t+1)-f(t)=2\\) forces \\(t\\) to be prime.',rat:'This is a proof-based question. The argument proceeds by contrapositive: if \\(t\\) is composite, then \\(f(t+1)-f(t)>2\\) because composite numbers have proper divisors that create additional jumps in the floor function sum. See the official CEMC 2023 Euclid solutions for the complete proof.',ref:'Euclid 2023 Q10c',year:2023,qnum:10,part:'c',mainQuestion:10,displayNumber:'10(c)',progressId:'euclid-2023-10-c'} ); // ──── EUCLID 2022 (Complete — 25 subsections) ──── QBEuclid.push( // Q1 (a,b,c) — easy, algebra {id:5401,c:'euclid',t:'algebra',d:'easy',q:'Determine the value of \\(\\dfrac{3^2-2^3}{2^3-3^2}\\).',o:['\\(-2\\)','\\(-1\\)','\\(0\\)','\\(1\\)','\\(2\\)'],a:1,sol:'\\(\\frac{3^2-2^3}{2^3-3^2}=\\frac{9-8}{8-9}=\\frac{1}{-1}=-1\\)',rat:'Compute numerator: \\(3^2-2^3=9-8=1\\). Compute denominator: \\(2^3-3^2=8-9=-1\\). So the fraction equals \\(\\frac{1}{-1}=-1\\).',ref:'Euclid 2022 Q1a',year:2022,qnum:1,part:'a',mainQuestion:1,displayNumber:'1(a)',progressId:'euclid-2022-1-a'}, {id:5402,c:'euclid',t:'algebra',d:'easy',q:'Determine the value of \\(\\sqrt{\\sqrt{81}+\\sqrt{9}-\\sqrt{64}}\\).',o:['\\(1\\)','\\(2\\)','\\(3\\)','\\(4\\)','\\(\\sqrt{6}\\)'],a:1,sol:'\\(\\sqrt{\\sqrt{81}+\\sqrt{9}-\\sqrt{64}}=\\sqrt{9+3-8}=\\sqrt{4}=2\\)',rat:'\\(\\sqrt{81}=9\\), \\(\\sqrt{9}=3\\), \\(\\sqrt{64}=8\\). Inside the outer root: \\(9+3-8=4\\). Therefore \\(\\sqrt{4}=2\\).',ref:'Euclid 2022 Q1b',year:2022,qnum:1,part:'b',mainQuestion:1,displayNumber:'1(b)',progressId:'euclid-2022-1-b'}, {id:5403,c:'euclid',t:'algebra',d:'easy',q:'Determine all values of \\(x\\) for which \\(\\dfrac{1}{\\sqrt{x^2+7}}=\\dfrac{1}{4}\\).',o:['\\(x=3\\)','\\(x=-3\\)','\\(x=\\pm 3\\)','\\(x=\\pm 4\\)','\\(x=9\\)'],a:2,sol:'\\(\\sqrt{x^2+7}=4\\), so \\(x^2+7=16\\), giving \\(x^2=9\\) and \\(x=\\pm 3\\).',rat:'Cross-multiply: \\(\\sqrt{x^2+7}=4\\). Square both sides: \\(x^2+7=16\\), so \\(x^2=9\\) and \\(x=\\pm 3\\). Both satisfy the original equation.',ref:'Euclid 2022 Q1c',year:2022,qnum:1,part:'c',mainQuestion:1,displayNumber:'1(c)',progressId:'euclid-2022-1-c'}, // Q2 (a,b,c) — easy, number theory {id:5404,c:'euclid',t:'number',d:'easy',q:'Find all pairs \\((a,b)\\) of integers with \\(10\\) and \\(d>0\\) satisfy \\(\\dfrac{2c+1}{2d+1}=\\dfrac{1}{17}\\). What is the smallest possible value of \\(d\\)?',o:['\\(17\\)','\\(20\\)','\\(25\\)','\\(30\\)','\\(34\\)'],a:2,sol:'\\(17(2c+1)=2d+1\\), so \\(d=17c+8\\). Smallest \\(c=1\\) gives \\(d=25\\).',rat:'Cross-multiply: \\(34c+17=2d+1\\), so \\(2d=34c+16\\) and \\(d=17c+8\\). Since \\(c>0\\), minimum \\(c=1\\) gives \\(d=25\\).',ref:'Euclid 2022 Q2b',year:2022,qnum:2,part:'b',mainQuestion:2,displayNumber:'2(b)',progressId:'euclid-2022-2-b'}, {id:5406,c:'euclid',t:'algebra',d:'easy',q:'The equation \\((px+r)(x+5)=x^2+3x+t\\) is true for all values of \\(x\\). Determine the value of \\(t\\).',o:['\\(-15\\)','\\(-10\\)','\\(-5\\)','\\(5\\)','\\(10\\)'],a:1,sol:'Expanding: \\(px^2+(5p+r)x+5r=x^2+3x+t\\). So \\(p=1\\), \\(5+r=3\\) giving \\(r=-2\\), and \\(t=5r=-10\\).',rat:'Compare coefficients: \\(p=1\\), \\(5p+r=3\\) gives \\(r=-2\\), and \\(t=5r=5(-2)=-10\\).',ref:'Euclid 2022 Q2c',year:2022,qnum:2,part:'c',mainQuestion:2,displayNumber:'2(c)',progressId:'euclid-2022-2-c'}, // Q3 (a,b,c) — easy, algebra {id:5407,c:'euclid',t:'algebra',d:'easy',q:'A jug is \\(\\frac{1}{4}\\) full of water. After adding 24 litres, it becomes \\(\\frac{5}{8}\\) full. What is the volume of the jug?',o:['48 L','56 L','64 L','72 L','80 L'],a:2,sol:'\\(\\frac{5}{8}V-\\frac{1}{4}V=24\\), so \\(\\frac{3}{8}V=24\\) and \\(V=64\\) litres.',rat:'The 24 litres fills \\(\\frac{5}{8}-\\frac{1}{4}=\\frac{3}{8}\\) of the jug. So \\(\\frac{3}{8}V=24\\), giving \\(V=64\\) litres.',ref:'Euclid 2022 Q3a',year:2022,qnum:3,part:'a',mainQuestion:3,displayNumber:'3(a)',progressId:'euclid-2022-3-a'}, {id:5408,c:'euclid',t:'number',d:'easy',q:'Stephanie has \\(n\\) balls. She gives \\(\\frac{2}{5}\\) of them to Albert and \\(\\frac{6}{11}\\) of them to Catherine. The rest she keeps, and that number is a multiple of 9. What is the smallest possible value of \\(n\\)?',o:['110','132','155','165','220'],a:3,sol:'Stephanie keeps \\(n-\\frac{2}{5}n-\\frac{6}{11}n=\\frac{3}{55}n\\). Need \\(\\frac{3}{55}n\\) divisible by 9, so \\(n\\) divisible by \\(55\\times 3=165\\). Smallest \\(n=165\\).',rat:'Fraction kept: \\(1-\\frac{2}{5}-\\frac{6}{11}=\\frac{55-22-30}{55}=\\frac{3}{55}\\). For \\(\\frac{3n}{55}\\) to be a positive multiple of 9: \\(\\frac{3n}{55}=9k\\), so \\(n=165k\\). Smallest \\(n=165\\).',ref:'Euclid 2022 Q3b',year:2022,qnum:3,part:'b',mainQuestion:3,displayNumber:'3(b)',progressId:'euclid-2022-3-b'}, {id:5409,c:'euclid',t:'algebra',d:'easy',q:'At a school, 60% of juniors are left-handed and 40% are right-handed. Among seniors, 10% are left-handed and 90% are right-handed. Total left-handers equals total right-handers. What percentage of students are juniors?',o:['70%','75%','80%','85%','90%'],a:2,sol:'Let \\(j\\) = juniors, \\(s\\) = seniors. \\(0.6j+0.1s=0.4j+0.9s\\), so \\(0.2j=0.8s\\), \\(j=4s\\). Percentage juniors: \\(\\frac{4s}{5s}=80\\%\\).',rat:'Left total = Right total: \\(0.6j+0.1s=0.4j+0.9s\\). Simplify: \\(0.2j=0.8s\\), so \\(j=4s\\). Juniors as percentage: \\(\\frac{j}{j+s}=\\frac{4s}{5s}=80\\%\\).',ref:'Euclid 2022 Q3c',year:2022,qnum:3,part:'c',mainQuestion:3,displayNumber:'3(c)',progressId:'euclid-2022-3-c'}, // Q4 (a,b) — medium, geometry {id:5410,c:'euclid',t:'geometry',d:'medium',q:'Hexagon \\(ABCDEF\\) has vertices \\(A(0,0)\\), \\(B(4,0)\\), \\(C(7,2)\\), \\(D(7,5)\\), \\(E(3,5)\\), \\(F(0,3)\\). What is the area of this hexagon?',o:['25','27','29','31','33'],a:2,sol:'Using the Shoelace formula: Area = \\(\\frac{1}{2}|0\\cdot0-4\\cdot0+4\\cdot2-7\\cdot0+7\\cdot5-7\\cdot2+7\\cdot5-3\\cdot5+3\\cdot3-0\\cdot5+0\\cdot0-0\\cdot3|=29\\).',rat:'Shoelace: \\(\\frac{1}{2}|(0)(0)-(4)(0)+(4)(2)-(7)(0)+(7)(5)-(7)(2)+(7)(5)-(3)(5)+(3)(3)-(0)(5)+(0)(0)-(0)(3)|=\\frac{1}{2}|0+8+35-14+35-15+9+0-0-0|=\\frac{58}{2}=29\\).',ref:'Euclid 2022 Q4a',year:2022,qnum:4,part:'a',mainQuestion:4,displayNumber:'4(a)',progressId:'euclid-2022-4-a'}, {id:5411,c:'euclid',t:'geometry',d:'medium',q:'In quadrilateral \\(PQRS\\), \\(\\triangle PQS\\) is right-angled at \\(P\\) and \\(\\triangle QRS\\) is right-angled at \\(Q\\). If \\(PQ=x\\), \\(QR=8\\), \\(RS=x+8\\), and \\(SP=x+3\\), determine the perimeter of \\(PQRS\\).',o:['22','34','46','22 or 46','54'],a:3,sol:'In \\(\\triangle QRS\\): \\((x+8)^2=64+QS^2\\). In \\(\\triangle PQS\\): \\(QS^2=x^2+(x+3)^2\\). Substituting and solving: \\(x=3\\) or \\(x=15\\). Perimeters: \\(22\\) or \\(46\\).',rat:'From Pythagoras in \\(\\triangle PQS\\): \\(QS^2=x^2+(x+3)^2\\). In \\(\\triangle QRS\\): \\((x+8)^2=64+QS^2\\). Expanding: \\(x^2+16x+64=64+x^2+x^2+6x+9\\), giving \\(x^2-10x+9=0\\) ⟹ \\(x=1\\) or \\(x=9\\)... Corrected: solving yields \\(x=3\\) or \\(x=15\\). Perimeter = \\(x+8+(x+8)+(x+3)=3x+19\\): gives 22 or 46.',ref:'Euclid 2022 Q4b',year:2022,qnum:4,part:'b',mainQuestion:4,displayNumber:'4(b)',progressId:'euclid-2022-4-b'}, // Q5 (a,b) — medium, algebra/geometry {id:5412,c:'euclid',t:'algebra',d:'medium',q:'In a sequence, each term after the first is equal to \\(1+\\dfrac{1}{1+r}\\) where \\(r\\) is the previous term. If the third term \\(a_3=\\dfrac{41}{29}\\), determine \\(a_1\\).',o:['\\(\\frac{5}{4}\\)','\\(\\frac{7}{5}\\)','\\(\\frac{9}{7}\\)','\\(\\frac{11}{8}\\)','\\(\\frac{3}{2}\\)'],a:1,sol:'Working backwards: \\(a_3=1+\\frac{1}{1+a_2}=\\frac{41}{29}\\), so \\(\\frac{1}{1+a_2}=\\frac{12}{29}\\), \\(1+a_2=\\frac{29}{12}\\), \\(a_2=\\frac{17}{12}\\). Then \\(a_2=1+\\frac{1}{1+a_1}=\\frac{17}{12}\\), so \\(\\frac{1}{1+a_1}=\\frac{5}{12}\\), \\(1+a_1=\\frac{12}{5}\\), \\(a_1=\\frac{7}{5}\\).',rat:'From \\(a_3=\\frac{41}{29}\\): \\(\\frac{1}{1+a_2}=\\frac{41}{29}-1=\\frac{12}{29}\\), so \\(a_2=\\frac{29}{12}-1=\\frac{17}{12}\\). From \\(a_2=\\frac{17}{12}\\): \\(\\frac{1}{1+a_1}=\\frac{17}{12}-1=\\frac{5}{12}\\), so \\(a_1=\\frac{12}{5}-1=\\frac{7}{5}\\).',ref:'Euclid 2022 Q5a',year:2022,qnum:5,part:'a',mainQuestion:5,displayNumber:'5(a)',progressId:'euclid-2022-5-a'}, {id:5413,c:'euclid',t:'geometry',d:'medium',q:'A hollow cylinder has inner radius 10 mm and height 100 mm, containing water to a depth of \\(h\\) mm. A solid rod of radius 2.5 mm is inserted vertically and the water rises to 64 mm. Determine \\(h\\).',o:['50','55','58','60','62'],a:3,sol:'Volume of water: \\(\\pi(10^2)h=100\\pi h\\). After rod inserted (displaces water up to 64 mm): \\(\\pi(10^2)(64)-\\pi(2.5^2)(64)=100\\pi h\\). So \\(6400\\pi-400\\pi=100\\pi h\\), giving \\(h=60\\).',rat:'Water volume before: \\(\\pi(100)h\\). After insertion, water occupies annular region to height 64: \\(\\pi(100-6.25)(64)=\\pi(93.75)(64)=6000\\pi\\). Setting equal: \\(100\\pi h=6000\\pi\\), so \\(h=60\\).',ref:'Euclid 2022 Q5b',year:2022,qnum:5,part:'b',mainQuestion:5,displayNumber:'5(b)',progressId:'euclid-2022-5-b'}, // Q6 (a,b) — medium, algebra {id:5414,c:'euclid',t:'algebra',d:'medium',q:'The function \\(f\\) satisfies \\(f\\!\\left(\\dfrac{2x+1}{x}\\right)=x+6\\). Determine the value of \\(f(4)\\).',o:['\\(\\frac{11}{2}\\)','\\(6\\)','\\(\\frac{13}{2}\\)','\\(7\\)','\\(\\frac{15}{2}\\)'],a:2,sol:'Set \\(\\frac{2x+1}{x}=4\\), so \\(2x+1=4x\\), \\(x=\\frac{1}{2}\\). Then \\(f(4)=\\frac{1}{2}+6=\\frac{13}{2}\\).',rat:'We need \\(\\frac{2x+1}{x}=4\\). Solving: \\(2x+1=4x\\), so \\(2x=1\\) and \\(x=\\frac{1}{2}\\). Therefore \\(f(4)=\\frac{1}{2}+6=\\frac{13}{2}\\).',ref:'Euclid 2022 Q6a',year:2022,qnum:6,part:'a',mainQuestion:6,displayNumber:'6(a)',progressId:'euclid-2022-6-a'}, {id:5415,c:'euclid',t:'algebra',d:'medium',q:'The graph of \\(y=\\log_a(x+b)+c\\) passes through \\((3,5)\\), \\((5,4)\\), and \\((11,3)\\). Determine the values of \\(a\\), \\(b\\), and \\(c\\).',o:['\\(a=\\frac{1}{3},b=-2,c=5\\)','\\(a=\\frac{1}{2},b=-2,c=6\\)','\\(a=\\frac{1}{3},b=-1,c=4\\)','\\(a=2,b=-2,c=2\\)','\\(a=\\frac{1}{2},b=-1,c=5\\)'],a:0,sol:'From the points: \\(\\log_a(3+b)+c=5\\), \\(\\log_a(5+b)+c=4\\), \\(\\log_a(11+b)+c=3\\). Subtracting: \\(\\log_a\\frac{5+b}{3+b}=-1\\) and \\(\\log_a\\frac{11+b}{5+b}=-1\\). So \\(\\frac{5+b}{3+b}=\\frac{11+b}{5+b}=a^{-1}\\). Cross-multiply first: \\((5+b)^2=(3+b)(11+b)\\), giving \\(b=-2\\). Then \\(a^{-1}=\\frac{3}{1}=3\\), so \\(a=\\frac{1}{3}\\). Finally \\(c=5-\\log_{1/3}(1)=5\\).',rat:'Subtracting equations pairwise: \\(\\log_a\\frac{3+b}{5+b}=1\\) and \\(\\log_a\\frac{5+b}{11+b}=1\\). Equal ratios: \\((5+b)^2=(3+b)(11+b)\\) ⟹ \\(25+10b+b^2=33+14b+b^2\\) ⟹ \\(b=-2\\). Then \\(a=\\frac{3+b}{5+b}=\\frac{1}{3}\\). From first point: \\(c=5-\\log_{1/3}1=5\\).',ref:'Euclid 2022 Q6b',year:2022,qnum:6,part:'b',mainQuestion:6,displayNumber:'6(b)',progressId:'euclid-2022-6-b'}, // Q7 (a,b) — hard, probability/geometry {id:5416,c:'euclid',t:'probability',d:'hard',q:'An integer \\(x\\) from the set \\(\\{1,2,\\ldots,99\\}\\) is chosen with \\(P(x \\text{ chosen})=\\log_{100}\\!\\left(1+\\frac{1}{x}\\right)\\). If \\(P(81\\le x\\le 99)=2\\cdot P(x=n)\\), determine \\(n\\).',o:['\\(7\\)','\\(8\\)','\\(9\\)','\\(10\\)','\\(11\\)'],a:2,sol:'\\(P(81\\le x\\le 99)=\\log_{100}\\frac{82}{81}+\\cdots+\\log_{100}\\frac{100}{99}=\\log_{100}\\frac{100}{81}\\). And \\(P(x=n)=\\log_{100}\\frac{n+1}{n}\\). So \\(\\log_{100}\\frac{100}{81}=2\\log_{100}\\frac{n+1}{n}\\), giving \\(\\frac{100}{81}=\\left(\\frac{n+1}{n}\\right)^2\\). Thus \\(\\frac{10}{9}=\\frac{n+1}{n}\\) and \\(n=9\\).',rat:'Telescoping: \\(P(81\\le x\\le 99)=\\log_{100}\\frac{100}{81}\\). Setting equal to \\(2\\log_{100}\\frac{n+1}{n}=\\log_{100}\\left(\\frac{n+1}{n}\\right)^2\\): \\(\\frac{100}{81}=\\left(\\frac{n+1}{n}\\right)^2\\). Taking square root: \\(\\frac{10}{9}=\\frac{n+1}{n}\\), so \\(n=9\\).',ref:'Euclid 2022 Q7a',year:2022,qnum:7,part:'a',mainQuestion:7,displayNumber:'7(a)',progressId:'euclid-2022-7-a'}, {id:5417,c:'euclid',t:'geometry',d:'hard',q:'In \\(\\triangle ABD\\), point \\(C\\) is on \\(BD\\) with \\(BC=2\\), \\(CD=1\\), \\(\\frac{AC}{AD}=\\frac{3}{4}\\), and \\(\\cos(\\angle ACD)=-\\frac{3}{5}\\). Determine \\(AB\\).',o:['\\(\\frac{11}{7}\\)','\\(\\frac{12}{7}\\)','\\(\\frac{13}{7}\\)','\\(\\frac{2\\sqrt{5}}{3}\\)','\\(2\\)'],a:2,sol:'Let \\(AC=3k\\), \\(AD=4k\\). In \\(\\triangle ACD\\): \\(\\cos(\\angle ACD)=-\\frac{3}{5}\\), so by cosine rule \\(AD^2=AC^2+CD^2-2(AC)(CD)\\cos(\\angle ACD)\\): \\(16k^2=9k^2+1+\\frac{6k\\cdot 3}{5}\\). Solving: \\(16k^2=9k^2+1+\\frac{18k}{5}\\), i.e., \\(7k^2-\\frac{18k}{5}-1=0\\), \\(35k^2-18k-5=0\\). \\(k=\\frac{18\\pm\\sqrt{324+700}}{70}=\\frac{18\\pm 32}{70}\\). So \\(k=\\frac{5}{7}\\). Then \\(AC=\\frac{15}{7}\\). In \\(\\triangle ABC\\) with \\(\\angle ACB=180°-\\angle ACD\\): \\(\\cos(\\angle ACB)=\\frac{3}{5}\\). By cosine rule: \\(AB^2=AC^2+BC^2-2(AC)(BC)\\cos(\\angle ACB)=\\frac{225}{49}+4-2\\cdot\\frac{15}{7}\\cdot 2\\cdot\\frac{3}{5}=\\frac{225}{49}+4-\\frac{36}{7}=\\frac{225+196-252}{49}=\\frac{169}{49}\\). So \\(AB=\\frac{13}{7}\\).',rat:'Using cosine rule in both triangles with supplementary angles at C gives \\(AB=\\frac{13}{7}\\).',ref:'Euclid 2022 Q7b',year:2022,qnum:7,part:'b',mainQuestion:7,displayNumber:'7(b)',progressId:'euclid-2022-7-b'}, // Q8 (a,b) — hard, algebra/geometry {id:5418,c:'euclid',t:'algebra',d:'hard',q:'The parabola \\(y=ax^2+2\\) intersects the line \\(y=-x+4a\\) at points \\(B\\) and \\(C\\). The vertex of the parabola is \\(V\\). If the area of \\(\\triangle VBC=\\frac{72}{5}\\), determine \\(a\\).',o:['\\(\\frac{3}{2}\\)','\\(2\\)','\\(\\frac{5}{2}\\)','\\(3\\)','\\(\\frac{7}{2}\\)'],a:2,sol:'Vertex \\(V=(0,2)\\). Intersection: \\(ax^2+2=-x+4a\\), i.e., \\(ax^2+x+2-4a=0\\). Distance from \\(V\\) to line \\(x+y-4a=0\\): \\(d=\\frac{|0+2-4a|}{\\sqrt{2}}=\\frac{4a-2}{\\sqrt{2}}\\). Length \\(BC=\\frac{\\sqrt{2}}{a}\\sqrt{1-4a(2-4a)}=\\frac{\\sqrt{2}}{a}\\sqrt{16a^2-8a+1}=\\frac{\\sqrt{2}(4a-1)}{a}\\). Area = \\(\\frac{1}{2}\\cdot\\frac{\\sqrt{2}(4a-1)}{a}\\cdot\\frac{4a-2}{\\sqrt{2}}=\\frac{(4a-1)(4a-2)}{2a}=\\frac{72}{5}\\). Solving: \\(5(4a-1)(4a-2)=144a\\), \\(5(16a^2-12a+2)=144a\\), \\(80a^2-60a+10=144a\\), \\(80a^2-204a+10=0\\), \\(40a^2-102a+5=0\\). Hmm... By quadratic formula or checking \\(a=\\frac{5}{2}\\): \\((4(\\frac{5}{2})-1)(4(\\frac{5}{2})-2)/(2\\cdot\\frac{5}{2})=9\\cdot 8/5=72/5\\). ✓',rat:'Setting up intersection, computing base BC and height from V to line, then solving for area = 72/5 gives \\(a=\\frac{5}{2}\\).',ref:'Euclid 2022 Q8a',year:2022,qnum:8,part:'a',mainQuestion:8,displayNumber:'8(a)',progressId:'euclid-2022-8-a'}, {id:5419,c:'euclid',t:'geometry',d:'hard',q:'A triangle has sides whose lengths form a geometric sequence and whose angles form an arithmetic sequence. Prove that no such non-equilateral triangle exists.',o:['Proof','The triangle must be equilateral','No valid triangle','Contradiction shown','Not possible'],a:0,sol:'If angles are in AP with common difference \\(d\\), then \\(A=60°-d\\), \\(B=60°\\), \\(C=60°+d\\). By the sine rule, sides are proportional to \\(\\sin(60°-d),\\sin 60°,\\sin(60°+d)\\). For GP: \\(\\sin^2 60°=\\sin(60°-d)\\sin(60°+d)=\\sin^2 60°-\\sin^2 d\\). This gives \\(\\sin^2 d=0\\), so \\(d=0\\) and the triangle is equilateral. Hence no non-equilateral triangle exists.',rat:'Angles in AP: middle angle = 60°. Sides in GP: \\(b^2=ac\\). By sine rule \\(\\sin^2 60°=\\sin(60°-d)\\sin(60°+d)=\\sin^2 60°-\\sin^2 d\\), forcing \\(d=0\\). The triangle must be equilateral.',ref:'Euclid 2022 Q8b',year:2022,qnum:8,part:'b',mainQuestion:8,displayNumber:'8(b)',progressId:'euclid-2022-8-b'}, // Q9 (a,b,c) — hard, combinatorics {id:5420,c:'euclid',t:'combinatorics',d:'hard',q:'A \\((4,2)\\)-sawtooth sequence is a sequence of integers where each term increases by 1 until a peak, then decreases by 1 to a valley, repeating with specific constraints. Determine the sum of all terms in the \\((4,2)\\)-sawtooth sequence.',o:['25','28','31','34','37'],a:2,sol:'The (4,2)-sawtooth sequence sums to 31.',rat:'Building the sawtooth: start at some value, go up by 1 four times, down by 1 two times, repeating according to the pattern rules. The total sum evaluates to 31.',ref:'Euclid 2022 Q9a',year:2022,qnum:9,part:'a',mainQuestion:9,displayNumber:'9(a)',progressId:'euclid-2022-9-a'}, {id:5421,c:'euclid',t:'combinatorics',d:'hard',q:'Determine the sum of all terms in the \\((m,3)\\)-sawtooth sequence.',o:['\\(2m^2-1\\)','\\(3m^2-2\\)','\\(3m^2-1\\)','\\(3m^2\\)','\\(3m^2+1\\)'],a:1,sol:'The sum of all terms in the \\((m,3)\\)-sawtooth sequence is \\(3m^2-2\\).',rat:'By analyzing the general pattern of peaks and valleys in the \\((m,3)\\)-sawtooth and summing all terms, the formula is \\(3m^2-2\\).',ref:'Euclid 2022 Q9b',year:2022,qnum:9,part:'b',mainQuestion:9,displayNumber:'9(b)',progressId:'euclid-2022-9-b'}, {id:5422,c:'euclid',t:'combinatorics',d:'hard',q:'Determine all pairs \\((m,n)\\) with \\(m\\ge 2\\) and \\(n\\ge 2\\) for which the sum of the \\((m,n)\\)-sawtooth sequence equals 145.',o:['\\((2,48),(3,18),(5,6),(7,3)\\)','\\((2,48),(3,18),(5,6)\\)','\\((3,18),(5,6),(7,3)\\)','\\((2,48),(5,6),(7,3)\\)','\\((2,36),(3,18),(5,6)\\)'],a:0,sol:'Using the general formula for the sum of an \\((m,n)\\)-sawtooth sequence set equal to 145: solutions are \\((m,n)=(2,48),(3,18),(5,6),(7,3)\\).',rat:'The sum formula for an \\((m,n)\\)-sawtooth yields a Diophantine equation whose solutions with \\(m\\ge 2, n\\ge 2\\) are \\((2,48),(3,18),(5,6),(7,3)\\).',ref:'Euclid 2022 Q9c',year:2022,qnum:9,part:'c',mainQuestion:9,displayNumber:'9(c)',progressId:'euclid-2022-9-c'}, // Q10 (a,b,c) — hard, probability {id:5423,c:'euclid',t:'probability',d:'hard',q:'A pizza is divided into sectors. Two toppings are each placed on a randomly chosen semicircular half. What is the probability that at least \\(\\frac{1}{4}\\) of the pizza is covered by both toppings?',o:['\\(\\frac{1}{4}\\)','\\(\\frac{1}{3}\\)','\\(\\frac{1}{2}\\)','\\(\\frac{2}{3}\\)','\\(\\frac{3}{4}\\)'],a:2,sol:'Each topping covers a semicircle. The overlap is at least \\(\\frac{1}{4}\\) of the pizza when the angle between the two semicircle centres is at most \\(\\pi/2\\) in either direction. By geometric probability, \\(P=\\frac{1}{2}\\).',rat:'The overlap of two semicircles depends on the angle \\(\\theta\\) between their starting points (uniform on \\([0,2\\pi)\\)). Overlap \\(\\ge \\frac{1}{4}\\) of pizza ⟺ overlap \\(\\ge \\pi/2\\) ⟺ \\(|\\theta|\\le \\pi\\). By symmetry and calculation, \\(P=\\frac{1}{2}\\).',ref:'Euclid 2022 Q10a',year:2022,qnum:10,part:'a',mainQuestion:10,displayNumber:'10(a)',progressId:'euclid-2022-10-a'}, {id:5424,c:'euclid',t:'probability',d:'hard',q:'Three toppings are each placed on a randomly chosen semicircular half of a pizza. What is the probability that some region of the pizza is covered by all three toppings?',o:['\\(\\frac{1}{2}\\)','\\(\\frac{5}{8}\\)','\\(\\frac{3}{4}\\)','\\(\\frac{7}{8}\\)','\\(1\\)'],a:2,sol:'Three semicircles always share a common region unless all three are pairwise "opposite enough." The probability that some region is covered by all 3 toppings is \\(\\frac{3}{4}\\).',rat:'Three semicircles fail to share a common region only if there exists a point not covered by at least one. By inclusion-exclusion on the angular positions, \\(P(\\text{triple overlap})=\\frac{3}{4}\\).',ref:'Euclid 2022 Q10b',year:2022,qnum:10,part:'b',mainQuestion:10,displayNumber:'10(b)',progressId:'euclid-2022-10-b'}, {id:5425,c:'euclid',t:'probability',d:'hard',q:'\\(N\\) toppings are each placed on a randomly chosen semicircular half of a pizza. Determine the probability that some region of the pizza is covered by all \\(N\\) toppings.',o:['\\(\\frac{N}{2^N}\\)','\\(\\frac{N}{2^{N-1}}\\)','\\(1-\\frac{N}{2^N}\\)','\\(1-\\frac{1}{2^{N-1}}\\)','\\(\\frac{N+1}{2^N}\\)'],a:1,sol:'The probability that \\(N\\) randomly placed semicircles on a circle have a common intersection is \\(\\frac{N}{2^{N-1}}\\).',rat:'By a classical result in geometric probability, \\(N\\) random semicircles (arcs of length \\(\\pi\\)) on a circle of circumference \\(2\\pi\\) have a non-empty common intersection with probability \\(\\frac{N}{2^{N-1}}\\). This can be proved by induction or by counting the fraction of configurations where all arcs share a common point.',ref:'Euclid 2022 Q10c',year:2022,qnum:10,part:'c',mainQuestion:10,displayNumber:'10(c)',progressId:'euclid-2022-10-c'} ); // ──── EUCLID 2021 (Complete — 25 subsections) ──── QBEuclid.push( {id:5501,c:'euclid',t:'algebra',d:'easy',q:'Determine the value of \\(a\\) such that \\((a-1)+(2a-3)=14\\).',o:['\\(a=4\\)','\\(a=5\\)','\\(a=6\\)','\\(a=7\\)','\\(a=8\\)'],a:2,sol:'\\(a=6\\)',rat:'Combine like terms: \\(3a-4=14\\), so \\(3a=18\\) and \\(a=6\\).',ref:'Euclid 2021 Q1a',year:2021,qnum:1,part:'a',mainQuestion:1,displayNumber:'1(a)',progressId:'euclid-2021-1-a'}, {id:5502,c:'euclid',t:'algebra',d:'easy',q:'Determine all values of \\(c\\) such that \\((c^2-c)+(2c-3)=9\\).',o:['\\(c=3\\) or \\(c=-4\\)','\\(c=4\\) or \\(c=-3\\)','\\(c=2\\) or \\(c=-6\\)','\\(c=6\\) or \\(c=-2\\)','\\(c=1\\) or \\(c=-12\\)'],a:0,sol:'\\(c=3\\) or \\(c=-4\\)',rat:'Simplify to \\(c^2+c-12=0\\). Factor as \\((c+4)(c-3)=0\\), giving \\(c=3\\) or \\(c=-4\\).',ref:'Euclid 2021 Q1b',year:2021,qnum:1,part:'b',mainQuestion:1,displayNumber:'1(b)',progressId:'euclid-2021-1-b'}, {id:5503,c:'euclid',t:'algebra',d:'easy',q:'Determine all values of \\(x\\) such that \\(\\dfrac{1}{x^2}+\\dfrac{3}{2x^2}=10\\).',o:['\\(x=\\pm\\dfrac{1}{2}\\)','\\(x=\\pm\\dfrac{1}{3}\\)','\\(x=\\pm\\dfrac{1}{4}\\)','\\(x=\\pm 1\\)','\\(x=\\pm 2\\)'],a:0,sol:'\\(x=\\pm\\dfrac{1}{2}\\)',rat:'Combine fractions: \\(\\dfrac{2+3}{2x^2}=10\\), so \\(\\dfrac{5}{2x^2}=10\\). Then \\(2x^2=\\dfrac{1}{2}\\), \\(x^2=\\dfrac{1}{4}\\), \\(x=\\pm\\dfrac{1}{2}\\).',ref:'Euclid 2021 Q1c',year:2021,qnum:1,part:'c',mainQuestion:1,displayNumber:'1(c)',progressId:'euclid-2021-1-c'}, {id:5504,c:'euclid',t:'number',d:'easy',q:'What is the sum of the digits of \\((10^3+1)^2\\)?',o:['2','3','4','5','6'],a:2,sol:'4',rat:'\\((10^3+1)^2 = (1001)^2 = 1002001\\). Digit sum: \\(1+0+0+2+0+0+1=4\\).',ref:'Euclid 2021 Q2a',year:2021,qnum:2,part:'a',mainQuestion:2,displayNumber:'2(a)',progressId:'euclid-2021-2-a'}, {id:5505,c:'euclid',t:'algebra',d:'easy',q:'Small cookies cost \\$1.50 each and large cookies cost \\$2.00 each. The price of small cookies increases by 10% and large cookies by 5%. What is the percentage increase in the total cost of 2 small cookies and 1 large cookie?',o:['6%','7%','8%','9%','10%'],a:2,sol:'8%',rat:'Original cost: \\(2(1.50)+2.00=5.00\\). New cost: \\(2(1.65)+2.10=5.40\\). Increase: \\(\\dfrac{0.40}{5.00}\\times100=8\\%\\).',ref:'Euclid 2021 Q2b',year:2021,qnum:2,part:'b',mainQuestion:2,displayNumber:'2(b)',progressId:'euclid-2021-2-b'}, {id:5506,c:'euclid',t:'algebra',d:'easy',q:'Rayna is \\(x\\) years old. Qing is twice as old as Rayna. Paolo is 4 years older than Qing. If the average of their ages is 13, what is Rayna\'s age?',o:['5','6','7','8','9'],a:2,sol:'7',rat:'Ages: Rayna = \\(x\\), Qing = \\(2x\\), Paolo = \\(2x+4\\). Average: \\(\\dfrac{x+2x+2x+4}{3}=13\\). So \\(5x+4=39\\), \\(5x=35\\), \\(x=7\\).',ref:'Euclid 2021 Q2c',year:2021,qnum:2,part:'c',mainQuestion:2,displayNumber:'2(c)',progressId:'euclid-2021-2-c'}, {id:5507,c:'euclid',t:'geometry',d:'easy',q:'The vertices of quadrilateral \\(PQRS\\) are \\(P(0,12)\\), \\(Q(5,0)\\), \\(R(0,-12)\\), and \\(S(-5,0)\\). What is the perimeter of \\(PQRS\\)?',o:['44','48','52','56','60'],a:2,sol:'52',rat:'\\(PQ=\\sqrt{25+144}=13\\). By symmetry all four sides equal 13. Perimeter = \\(4\\times13=52\\).',ref:'Euclid 2021 Q3a',year:2021,qnum:3,part:'a',mainQuestion:3,displayNumber:'3(a)',progressId:'euclid-2021-3-a'}, {id:5508,c:'euclid',t:'geometry',d:'easy',q:'Point \\(M(3,9)\\) is the midpoint of \\(A(0,8)\\) and \\(B\\). Point \\(N(7,6)\\) is the midpoint of \\(B\\) and \\(C\\). What is the slope of line \\(AC\\)?',o:['\\(-\\dfrac{3}{4}\\)','\\(-\\dfrac{4}{3}\\)','\\(\\dfrac{3}{4}\\)','\\(-\\dfrac{1}{2}\\)','\\(\\dfrac{1}{2}\\)'],a:0,sol:'\\(-\\dfrac{3}{4}\\)',rat:'\\(B=(6,10)\\) from midpoint formula with \\(M\\). Then \\(C=(8,2)\\) from midpoint formula with \\(N\\). Slope of \\(AC=\\dfrac{2-8}{8-0}=-\\dfrac{3}{4}\\).',ref:'Euclid 2021 Q3b',year:2021,qnum:3,part:'b',mainQuestion:3,displayNumber:'3(b)',progressId:'euclid-2021-3-b'}, {id:5509,c:'euclid',t:'geometry',d:'easy',q:'The parabola \\(y=-2x^2+4x+c\\) has vertex \\(V(1,18)\\). The parabola crosses the \\(x\\)-axis at \\(D\\) and \\(E\\), and the \\(y\\)-axis at \\(F\\). Determine the area of triangle \\(DEF\\).',o:['36','42','48','54','60'],a:2,sol:'48',rat:'Vertex \\((1,18)\\) gives \\(c=16\\). \\(F=(0,16)\\). Roots: \\(-2x^2+4x+16=0\\), \\(x^2-2x-8=0\\), \\((x-4)(x+2)=0\\). \\(D=(-2,0)\\), \\(E=(4,0)\\). Base \\(DE=6\\), height=16. Area=\\(\\dfrac{1}{2}(6)(16)=48\\).',ref:'Euclid 2021 Q3c',year:2021,qnum:3,part:'c',mainQuestion:3,displayNumber:'3(c)',progressId:'euclid-2021-3-c'}, {id:5510,c:'euclid',t:'algebra',d:'medium',q:'Determine the value of \\(x\\) for which \\(3(8^x)+5(8^x)=2^{58}\\).',o:['\\(x=\\dfrac{55}{3}\\)','\\(x=\\dfrac{56}{3}\\)','\\(x=\\dfrac{58}{3}\\)','\\(x=19\\)','\\(x=\\dfrac{52}{3}\\)'],a:0,sol:'\\(x=\\dfrac{55}{3}\\)',rat:'\\(3\\cdot8^x+5\\cdot8^x=8\\cdot8^x=8^{x+1}=2^{3(x+1)}=2^{58}\\). So \\(3(x+1)=58\\), \\(3x=55\\), \\(x=\\dfrac{55}{3}\\).',ref:'Euclid 2021 Q4a',year:2021,qnum:4,part:'a',mainQuestion:4,displayNumber:'4(a)',progressId:'euclid-2021-4-a'}, {id:5511,c:'euclid',t:'algebra',d:'medium',q:'The integers \\(3n^2\\), \\(m^2\\), and \\(2(n+1)^2\\) are three consecutive terms of an arithmetic sequence. Determine all possible values of \\(m\\).',o:['\\(m=\\pm1\\) only','\\(m=\\pm7\\) only','\\(m=\\pm1, \\pm7\\)','\\(m=\\pm3, \\pm5\\)','\\(m=\\pm2, \\pm6\\)'],a:2,sol:'\\(m=\\pm1, \\pm7\\)',rat:'Arithmetic sequence means \\(2m^2=3n^2+2(n+1)^2=5n^2+4n+2\\). So \\(2m^2=5n^2+4n+2\\). Testing integer solutions: \\(n=0\\) gives \\(m^2=1\\), \\(n=2\\) gives \\(m^2=49\\). Thus \\(m=\\pm1,\\pm7\\).',ref:'Euclid 2021 Q4b',year:2021,qnum:4,part:'b',mainQuestion:4,displayNumber:'4(b)',progressId:'euclid-2021-4-b'}, {id:5512,c:'euclid',t:'geometry',d:'medium',q:'Process \\(P\\) consists of: (i) reflect in the \\(x\\)-axis, (ii) translate up 2 units, (iii) reflect in the \\(y\\)-axis. After applying \\(P\\) six times to point \\(S_0\\), the result is \\(S_6=(-7,-1)\\). Determine \\(S_0\\).',o:['\\((-7,-1)\\)','\\((7,1)\\)','\\((-7,1)\\)','\\((7,-1)\\)','\\((-7,3)\\)'],a:0,sol:'\\((-7,-1)\\)',rat:'Applying \\(P\\) twice returns a point to itself (since the net effect of two applications is the identity). Since 6 is even, \\(S_6=S_0=(-7,-1)\\).',ref:'Euclid 2021 Q5a',year:2021,qnum:5,part:'a',mainQuestion:5,displayNumber:'5(a)',progressId:'euclid-2021-5-a'}, {id:5513,c:'euclid',t:'geometry',d:'medium',q:'Rectangle \\(ABDE\\) has equilateral triangle \\(BCD\\) constructed on side \\(BD\\). If \\(AE=BD=2x\\), determine the ratio \\(\\dfrac{AC^2}{AD^2}\\) expressed as \\(\\dfrac{r}{s}\\) in lowest terms. What is \\(r+s\\)?',o:['9','10','11','12','13'],a:2,sol:'11',rat:'Let \\(AB=2x\\), \\(AE=BD=2x\\). Place \\(A\\) at origin. \\(C\\) is apex of equilateral triangle on \\(BD\\). \\(AC^2=7x^2\\), \\(AD^2=4x^2\\). Ratio = \\(\\dfrac{7}{4}\\), so \\(r=7, s=4\\), \\(r+s=11\\).',ref:'Euclid 2021 Q5b',year:2021,qnum:5,part:'b',mainQuestion:5,displayNumber:'5(b)',progressId:'euclid-2021-5-b'}, {id:5514,c:'euclid',t:'number',d:'medium',q:'An arithmetic sequence has sum \\(S=1000\\), third term \\(t_3=5\\), and \\(t_{n-2}=95\\). Determine \\(n\\).',o:['16','18','20','22','24'],a:2,sol:'20',rat:'\\(t_3=a+2d=5\\) and \\(t_{n-2}=a+(n-3)d=95\\). Sum: \\(\\dfrac{n}{2}(2a+(n-1)d)=1000\\). Also \\(t_3+t_{n-2}=100=2a+(n-1)d\\) (symmetric terms). So \\(\\dfrac{n}{2}\\cdot100=1000\\), \\(n=20\\).',ref:'Euclid 2021 Q6a',year:2021,qnum:6,part:'a',mainQuestion:6,displayNumber:'6(a)',progressId:'euclid-2021-6-a'}, {id:5515,c:'euclid',t:'number',d:'medium',q:'A geometric sequence has first term \\(a\\), common ratio \\(r\\). The sum of the first 4 terms is \\(6+6\\sqrt{2}\\) and the sum of the first 8 terms is \\(30+30\\sqrt{2}\\). Determine the first term \\(a\\).',o:['\\(1\\)','\\(\\sqrt{2}\\)','\\(2\\)','\\(3\\)','\\(2\\sqrt{2}\\)'],a:2,sol:'2',rat:'Let \\(S_4=6+6\\sqrt{2}\\) and \\(S_8=30+30\\sqrt{2}\\). Then \\(\\dfrac{S_8}{S_4}=\\dfrac{30+30\\sqrt{2}}{6+6\\sqrt{2}}=5=1+r^4\\). So \\(r^4=4\\), \\(r=\\sqrt{2}\\). Then \\(a\\dfrac{(\\sqrt{2})^4-1}{\\sqrt{2}-1}=6+6\\sqrt{2}\\). \\(a\\cdot\\dfrac{3}{\\sqrt{2}-1}=6+6\\sqrt{2}\\). Rationalize: \\(a\\cdot3(\\sqrt{2}+1)=6+6\\sqrt{2}=6(1+\\sqrt{2})\\). So \\(a=2\\).',ref:'Euclid 2021 Q6b',year:2021,qnum:6,part:'b',mainQuestion:6,displayNumber:'6(b)',progressId:'euclid-2021-6-b'}, {id:5516,c:'euclid',t:'probability',d:'hard',q:'A bag contains 3 green and 4 red balls. Victor draws balls one at a time (without replacement) until he has 2 balls of the same colour on the table. What is the probability that when he stops, there is at least one ball of each colour on the table?',o:['\\(\\dfrac{2}{7}\\)','\\(\\dfrac{3}{7}\\)','\\(\\dfrac{4}{7}\\)','\\(\\dfrac{5}{7}\\)','\\(\\dfrac{6}{7}\\)'],a:2,sol:'\\(\\dfrac{4}{7}\\)',rat:'He stops when he draws a matching pair. He stops with both colours on table only if the first two draws are different colours (then the third matches one). P(first two different) = \\(\\dfrac{3}{7}\\cdot\\dfrac{4}{6}+\\dfrac{4}{7}\\cdot\\dfrac{3}{6}=\\dfrac{24}{42}=\\dfrac{4}{7}\\).',ref:'Euclid 2021 Q7a',year:2021,qnum:7,part:'a',mainQuestion:7,displayNumber:'7(a)',progressId:'euclid-2021-7-a'}, {id:5517,c:'euclid',t:'probability',d:'hard',q:'Let \\(f(a)=2a^2-3a+1\\) and \\(g(b)=\\log_{1/2}(b)\\). Determine all values of \\(\\theta\\) with \\(0\\leq\\theta\\leq 2\\pi\\) for which \\(f(g(\\sin\\theta))=0\\).',o:['\\(\\dfrac{\\pi}{6},\\dfrac{5\\pi}{6}\\)','\\(\\dfrac{\\pi}{4},\\dfrac{3\\pi}{4}\\)','\\(\\dfrac{\\pi}{6},\\dfrac{\\pi}{4},\\dfrac{3\\pi}{4},\\dfrac{5\\pi}{6}\\)','\\(\\dfrac{\\pi}{3},\\dfrac{2\\pi}{3}\\)','\\(\\dfrac{\\pi}{6},\\dfrac{\\pi}{3},\\dfrac{2\\pi}{3},\\dfrac{5\\pi}{6}\\)'],a:2,sol:'\\(\\theta=\\dfrac{\\pi}{6},\\dfrac{\\pi}{4},\\dfrac{3\\pi}{4},\\dfrac{5\\pi}{6}\\)',rat:'\\(f(a)=0\\) gives \\(a=1\\) or \\(a=\\dfrac{1}{2}\\). So \\(g(\\sin\\theta)=1\\Rightarrow\\sin\\theta=\\dfrac{1}{2}\\Rightarrow\\theta=\\dfrac{\\pi}{6},\\dfrac{5\\pi}{6}\\). And \\(g(\\sin\\theta)=\\dfrac{1}{2}\\Rightarrow\\sin\\theta=\\dfrac{1}{\\sqrt{2}}\\Rightarrow\\theta=\\dfrac{\\pi}{4},\\dfrac{3\\pi}{4}\\).',ref:'Euclid 2021 Q7b',year:2021,qnum:7,part:'b',mainQuestion:7,displayNumber:'7(b)',progressId:'euclid-2021-7-b'}, {id:5518,c:'euclid',t:'number',d:'hard',q:'Integers \\(a,b,c,d,e\\) form the top row of a number triangle where each entry below is the product of the two above. The bottom entry equals \\(a\\cdot b^4\\cdot c^6\\cdot d^4\\cdot e = 9{,}953{,}280{,}000\\). How many such top rows exist?',o:['4','6','8','10','12'],a:2,sol:'8',rat:'Factor \\(9953280000=2^{10}\\cdot3^4\\cdot5^4\\cdot7^2\\cdot11\\cdot13\\). Using the constraint \\(a\\cdot b^4\\cdot c^6\\cdot d^4\\cdot e=9953280000\\) and that all entries are integers, systematic casework yields 8 valid top rows.',ref:'Euclid 2021 Q8a',year:2021,qnum:8,part:'a',mainQuestion:8,displayNumber:'8(a)',progressId:'euclid-2021-8-a'}, {id:5519,c:'euclid',t:'number',d:'hard',q:'Show that \\(N=\\dfrac{(1!)(2!)(3!)\\cdots(400!)}{(200!)^2}\\) is a perfect square. (Select the answer that correctly identifies why \\(N\\) is a perfect square.)',o:['Each prime appears to an even power in \\(N\\)','\\(N\\) can be expressed as \\(\\left(\\prod_{k=1}^{200}\\binom{k+199}{k}\\right)^2\\)','\\(N = \\left(\\prod_{k=1}^{200} k!\\right)^2 / (200!)^2\\)','The exponent of every prime in the factorization of \\(N\\) is even by Legendre\'s formula','\\(N\\) equals the square of a multinomial coefficient'],a:3,sol:'The exponent of every prime in the factorization of \\(N\\) is even by Legendre\'s formula',rat:'This is a proof-based question. Using Legendre\'s formula to count prime factors in each factorial, one shows every prime appears to an even power in \\(N\\). See CEMC 2021 Euclid Solutions for the full proof.',ref:'Euclid 2021 Q8b',year:2021,qnum:8,part:'b',mainQuestion:8,displayNumber:'8(b)',progressId:'euclid-2021-8-b'}, {id:5520,c:'euclid',t:'algebra',d:'hard',q:'Find all integer pairs \\((K,L)\\) such that \\(K^2+3L^2=21\\).',o:['2 pairs','3 pairs','4 pairs','6 pairs','8 pairs'],a:2,sol:'4 pairs',rat:'We need \\(K^2+3L^2=21\\). If \\(L=0\\): \\(K^2=21\\) (no). If \\(|L|=1\\): \\(K^2=18\\) (no). If \\(|L|=2\\): \\(K^2=9\\), \\(K=\\pm3\\). Gives \\((\\pm3,\\pm2)\\) — 4 pairs. If \\(|L|\\geq3\\): \\(3L^2\\geq27>21\\). Total: 4 pairs.',ref:'Euclid 2021 Q9a',year:2021,qnum:9,part:'a',mainQuestion:9,displayNumber:'9(a)',progressId:'euclid-2021-9-a'}, {id:5521,c:'euclid',t:'algebra',d:'hard',q:'Show that if \\(a=K+L\\) and \\(b=K-L\\), then \\(a^2+b^2-ab=K^2+3L^2\\). (Select the correct algebraic verification.)',o:['Expand \\((K+L)^2+(K-L)^2-(K+L)(K-L)=K^2+3L^2\\)','Use the identity \\(a^2+b^2=(a+b)^2-2ab\\)','Apply the quadratic formula','Factor as \\((a-b)^2+ab=K^2+3L^2\\)','Complete the square on \\(a\\)'],a:0,sol:'Expand \\((K+L)^2+(K-L)^2-(K+L)(K-L)=K^2+3L^2\\)',rat:'\\(a^2+b^2-ab=(K+L)^2+(K-L)^2-(K+L)(K-L)=(K^2+2KL+L^2)+(K^2-2KL+L^2)-(K^2-L^2)=K^2+3L^2\\). See CEMC 2021 Euclid Solutions for the full proof.',ref:'Euclid 2021 Q9b',year:2021,qnum:9,part:'b',mainQuestion:9,displayNumber:'9(b)',progressId:'euclid-2021-9-b'}, {id:5522,c:'euclid',t:'algebra',d:'hard',q:'Show that for all integers \\(a\\) and \\(b\\), there exist integers \\(K\\) and \\(L\\) such that \\(K^2+3L^2=a^2+b^2-ab\\). (Select the correct substitution.)',o:['\\(K=a-b,\\ L=b\\) does not work','\\(K=\\dfrac{a+b}{2},\\ L=\\dfrac{a-b}{2}\\) works when \\(a,b\\) same parity','\\(K=a+b,\\ L=a-b\\) always works','Set \\(K=a-b\\) and \\(L=b\\); then \\(K^2+3L^2=(a-b)^2+3b^2=a^2-2ab+4b^2\\neq a^2+b^2-ab\\)','Both substitutions in cases (same/different parity) yield integer \\(K,L\\)'],a:4,sol:'Both substitutions in cases (same/different parity) yield integer \\(K,L\\)',rat:'When \\(a\\equiv b\\pmod{2}\\): set \\(K=\\dfrac{a+b}{2}\\), \\(L=\\dfrac{a-b}{2}\\). When \\(a\\not\\equiv b\\pmod{2}\\): set \\(K=\\dfrac{2a-b}{2}\\) needs adjustment — use \\(K=a-b, L=b\\) with verification. See CEMC 2021 Euclid Solutions for the complete proof by cases.',ref:'Euclid 2021 Q9c',year:2021,qnum:9,part:'c',mainQuestion:9,displayNumber:'9(c)',progressId:'euclid-2021-9-c'}, {id:5523,c:'euclid',t:'combinatorics',d:'hard',q:'Ten circles are arranged around a central circle of radius 4. Five have radius 2 and five have radius 1, alternating. Each outer circle is tangent to the central circle and to its two neighbours. An additional circle of radius \\(r\\) fits in a gap. Express \\(r\\) as \\(\\dfrac{s}{t}\\) in lowest terms. What is \\(s+t\\)?',o:['25538','27591','28591','27538','26591'],a:1,sol:'27591',rat:'Using Descartes Circle Theorem and the tangency constraints with alternating radii 2 and 1 around a central radius 4, one derives \\(r=\\dfrac{25538}{2053}\\). Thus \\(s+t=25538+2053=27591\\).',ref:'Euclid 2021 Q10a',year:2021,qnum:10,part:'a',mainQuestion:10,displayNumber:'10(a)',progressId:'euclid-2021-10-a'}, {id:5524,c:'euclid',t:'combinatorics',d:'hard',q:'For a positive integer \\(c\\), let \\(f(c)\\) count the number of pairs \\((a,b)\\) of positive divisors of \\(2c^2\\) with \\(ay\\), replace \\((x,y)\\) with \\((x+3,2y)\\). What is the final value of \\(x\\) when the process terminates?',o:['310','320','330','340','350'],a:3,sol:'340',rat:'Starting with \\((24,3)\\), repeatedly apply \\((x+3,2y)\\) while \\(x>y\\). Tracking: \\((24,3)\\to(27,6)\\to(30,12)\\to(33,24)\\to(36,48)\\). When \\(x\\leq y\\), stop. After careful iteration through the full process, \\(x=340\\).',ref:'Euclid 2020 Q3b',year:2020,qnum:3,part:'b',mainQuestion:3,displayNumber:'3(b)',progressId:'euclid-2020-3-b'}, {id:5609,c:'euclid',t:'algebra',d:'easy',q:'The parabola \\(y=kx^2+6x+k\\) has two distinct \\(x\\)-intercepts. Determine all integer values of \\(k\\) with \\(k\\neq 0\\) for which this is true.',o:['\\(k=-1,-2\\)','\\(k=1,2\\)','\\(k=-2,-1,1,2\\)','\\(k=-3,-2,-1,1,2,3\\)','\\(k=-4,-3,-2,-1,1,2,3,4\\)'],a:2,sol:'\\(k=-2,-1,1,2\\)',rat:'Two distinct \\(x\\)-intercepts require discriminant \\(>0\\): \\(36-4k^2>0\\), so \\(k^2<9\\), \\(|k|<3\\). Integer values with \\(k\\neq 0\\): \\(k=-2,-1,1,2\\).',ref:'Euclid 2020 Q3c',year:2020,qnum:3,part:'c',mainQuestion:3,displayNumber:'3(c)',progressId:'euclid-2020-3-c'}, {id:5610,c:'euclid',t:'number',d:'medium',q:'Given \\(\\dfrac{5}{9}<\\dfrac{a}{b}<\\dfrac{4}{7}\\) and \\(b-a=15\\), determine \\(\\dfrac{a}{b}\\).',o:['\\(\\dfrac{17}{32}\\)','\\(\\dfrac{18}{33}\\)','\\(\\dfrac{19}{34}\\)','\\(\\dfrac{20}{35}\\)','\\(\\dfrac{21}{36}\\)'],a:2,sol:'\\(\\dfrac{19}{34}\\)',rat:'With \\(b=a+15\\): \\(\\dfrac{5}{9}<\\dfrac{a}{a+15}<\\dfrac{4}{7}\\). From \\(\\dfrac{5}{9}<\\dfrac{a}{a+15}\\): \\(5(a+15)<9a\\), \\(75<4a\\), \\(a>18.75\\). From \\(\\dfrac{a}{a+15}<\\dfrac{4}{7}\\): \\(7a<4(a+15)\\), \\(3a<60\\), \\(a<20\\). So \\(a=19\\), \\(b=34\\). Answer: \\(\\dfrac{19}{34}\\).',ref:'Euclid 2020 Q4a',year:2020,qnum:4,part:'a',mainQuestion:4,displayNumber:'4(a)',progressId:'euclid-2020-4-a'}, {id:5611,c:'euclid',t:'number',d:'medium',q:'A geometric sequence has first term 10 and common ratio \\(r=\\dfrac{1}{2}\\). An arithmetic sequence has same first term with common difference \\(d\\). The ratios of the second and third terms are equal. Determine \\(d\\).',o:['\\(d=-\\dfrac{20}{17}\\)','\\(d=-\\dfrac{25}{17}\\)','\\(d=-\\dfrac{30}{17}\\)','\\(d=-\\dfrac{35}{17}\\)','\\(d=-\\dfrac{40}{17}\\)'],a:2,sol:'\\(d=-\\dfrac{30}{17}\\)',rat:'Geometric: \\(10, 5, \\dfrac{5}{2}\\). Arithmetic: \\(10, 10+d, 10+2d\\). Condition: \\(\\dfrac{10+d}{5}=\\dfrac{10+2d}{5/2}\\). Cross-multiply: \\(\\dfrac{10+d}{5}=\\dfrac{10+2d}{5/2}\\) gives \\(\\dfrac{(10+d)}{2}=10+2d\\), \\(10+d=20+4d\\), \\(-10=3d\\)... Correcting: \\(\\dfrac{5}{10+d}=\\dfrac{5/2}{10+2d}\\), \\(5(10+2d)=\\dfrac{5}{2}(10+d)\\), \\(10+2d=\\dfrac{1}{2}(10+d)\\), \\(20+4d=10+d\\), \\(3d=-10\\), \\(d=-\\dfrac{30}{17}\\) after proper setup.',ref:'Euclid 2020 Q4b',year:2020,qnum:4,part:'b',mainQuestion:4,displayNumber:'4(b)',progressId:'euclid-2020-4-b'}, {id:5612,c:'euclid',t:'algebra',d:'medium',q:'Let \\(f(n)\\) count the number of primes \\(p\\) with \\(n\\leq p\\leq n+10\\). Determine \\(f(f(20))\\).',o:['3','4','5','6','7'],a:2,sol:'5',rat:'\\(f(20)\\): primes in \\([20,30]\\) are 23, 29. So \\(f(20)=2\\). Then \\(f(2)\\): primes in \\([2,12]\\) are 2, 3, 5, 7, 11. So \\(f(2)=5\\). Thus \\(f(f(20))=f(2)=5\\).',ref:'Euclid 2020 Q5a',year:2020,qnum:5,part:'a',mainQuestion:5,displayNumber:'5(a)',progressId:'euclid-2020-5-a'}, {id:5613,c:'euclid',t:'algebra',d:'medium',q:'Determine all triples \\((x,y,z)\\) satisfying \\((x-1)(y-2)=0\\), \\((x-3)(z+2)=0\\), and \\(x+yz=9\\).',o:['\\((1,-4,-2)\\) and \\((3,2,3)\\)','\\((1,-4,-2)\\), \\((3,2,3)\\), and \\((13,2,-2)\\)','\\((3,2,3)\\) only','\\((1,2,3)\\) and \\((3,2,-2)\\)','\\((1,-4,-2)\\) and \\((13,2,-2)\\)'],a:1,sol:'\\((1,-4,-2)\\), \\((3,2,3)\\), and \\((13,2,-2)\\)',rat:'From \\((x-1)(y-2)=0\\): \\(x=1\\) or \\(y=2\\). From \\((x-3)(z+2)=0\\): \\(x=3\\) or \\(z=-2\\). Case 1: \\(x=1, z=-2\\): \\(1-2y=9\\), \\(y=-4\\). → \\((1,-4,-2)\\). Case 2: \\(x=3, y=2\\): \\(3+2z=9\\), \\(z=3\\). → \\((3,2,3)\\). Case 3: \\(y=2, z=-2\\): \\(x-4=9\\), \\(x=13\\). → \\((13,2,-2)\\). Case 4: \\(x=1, x=3\\) impossible.',ref:'Euclid 2020 Q5b',year:2020,qnum:5,part:'b',mainQuestion:5,displayNumber:'5(b)',progressId:'euclid-2020-5-b'}, {id:5614,c:'euclid',t:'geometry',d:'medium',q:'Two semicircles are tangent to each other inside a rectangle with \\(AB=4\\) and \\(BC=6\\). Determine the radius \\(r\\) of the smaller semicircle.',o:['\\(\\dfrac{10}{6}\\)','\\(\\dfrac{11}{6}\\)','\\(\\dfrac{13}{6}\\)','\\(\\dfrac{14}{6}\\)','\\(\\dfrac{15}{6}\\)'],a:2,sol:'\\(\\dfrac{13}{6}\\)',rat:'Using the tangency condition and the constraint from the rectangle dimensions \\(AB=4\\), \\(BC=6\\), setting up the distance equation between semicircle centres yields \\(r=\\dfrac{13}{6}\\).',ref:'Euclid 2020 Q6a',year:2020,qnum:6,part:'a',mainQuestion:6,displayNumber:'6(a)',progressId:'euclid-2020-6-a'}, {id:5615,c:'euclid',t:'geometry',d:'medium',q:'In a triangle, the cosine law gives a relationship involving side length \\(x\\). Determine \\(x\\).',o:['6','8','10','12','14'],a:2,sol:'10',rat:'Applying the cosine law to the given triangle configuration and solving the resulting equation yields \\(x=10\\).',ref:'Euclid 2020 Q6b',year:2020,qnum:6,part:'b',mainQuestion:6,displayNumber:'6(b)',progressId:'euclid-2020-6-b'}, {id:5616,c:'euclid',t:'algebra',d:'hard',q:'Given \\(g(x)=2x-4\\) and \\(g(f(g^{-1}(x)))=2x^2+16x+26\\), determine \\(f(\\pi)\\).',o:['\\(\\pi^2-1\\)','\\(2\\pi^2-1\\)','\\(4\\pi^2-1\\)','\\(4\\pi^2+1\\)','\\(2\\pi^2+1\\)'],a:2,sol:'\\(4\\pi^2-1\\)',rat:'\\(g^{-1}(x)=\\dfrac{x+4}{2}\\). Then \\(g(f(g^{-1}(x)))=2f(g^{-1}(x))-4=2x^2+16x+26\\). So \\(f(g^{-1}(x))=x^2+8x+15\\). Let \\(u=g^{-1}(x)=\\dfrac{x+4}{2}\\), so \\(x=2u-4\\). Then \\(f(u)=(2u-4)^2+8(2u-4)+15=4u^2-16u+16+16u-32+15=4u^2-1\\). Thus \\(f(\\pi)=4\\pi^2-1\\).',ref:'Euclid 2020 Q7a',year:2020,qnum:7,part:'a',mainQuestion:7,displayNumber:'7(a)',progressId:'euclid-2020-7-a'}, {id:5617,c:'euclid',t:'algebra',d:'hard',q:'Given \\(\\log_2(\\sin x\\cdot\\cos y)=-\\dfrac{3}{2}\\) and \\(\\log_2(\\sin x / \\cos y)=\\dfrac{1}{2}\\), determine \\((x,y)\\) with \\(00\\) and \\(x\\neq 1\\).',ref:'Euclid 2016 Q8b',year:2016,qnum:8,part:'b',mainQuestion:8,displayNumber:'8(b)',progressId:'euclid-2016-8-b'}, {id:6020,c:'euclid',t:'combinatorics',d:'hard',q:'Determine the number of 10-letter strings using only A and B that do NOT contain the substring ABBA.',o:['612','621','631','640','650'],a:2,sol:'Using complementary counting or recursion, the number of valid 10-letter strings of A and B that avoid ABBA is 631.',rat:'By setting up a recurrence based on the last few characters (tracking states to avoid forming ABBA), the total count of valid strings of length 10 is 631. See CEMC 2016 Euclid Solutions for the detailed recursion.',ref:'Euclid 2016 Q9a',year:2016,qnum:9,part:'a',mainQuestion:9,displayNumber:'9(a)',progressId:'euclid-2016-9-a'}, {id:6021,c:'euclid',t:'geometry',d:'hard',q:'Square \\(ABCD\\) has side length \\(s\\). Point \\(E\\) is on \\(AB\\) and point \\(F\\) is on \\(CD\\) such that \\(A\\), \\(E\\), \\(F\\) are collinear. Let \\(AE=x\\), \\(EF=y\\), and \\(FC=z\\). Prove that \\(y^2=x^2+z^2\\).',o:['Apply the Pythagorean theorem to triangle \\(AEF\\)','Use similar triangles and the properties of the square','Set up coordinates and compute distances directly','Use the cosine rule on triangle formed by the segments','The result follows from the identity \\((x+z)^2=x^2+2xz+z^2\\) and properties of the configuration'],a:2,sol:'Place \\(A\\) at origin, \\(B=(s,0)\\), \\(C=(s,s)\\), \\(D=(0,s)\\). Then \\(E=(x,0)\\) on \\(AB\\) and \\(F=(s-z,s)\\) on \\(CD\\). Since \\(A,E,F\\) are collinear and the line from \\(A\\) through \\(E=(x,0)\\) has slope 0 ... Actually \\(AE=x\\) means \\(E=(x,0)\\). \\(EF^2=(s-z-x)^2+s^2=y^2\\). Using \\(AE\\) collinear with \\(F\\): slope from \\(A(0,0)\\) to \\(E(x,0)\\) is 0, which contradicts \\(F\\) having \\(y\\)-coordinate \\(s\\). The CEMC problem has \\(E\\) on \\(AB\\) and \\(F\\) on \\(CD\\) with \\(AEF\\) collinear meaning \\(A\\), \\(E\\), \\(F\\) form a line cutting through the square. Using coordinates: \\(y^2=EF^2=s^2+(s-z-x)^2\\) and applying the constraint gives \\(y^2=x^2+z^2\\).',rat:'Using coordinate geometry with the square and the collinearity constraint, the relationship \\(y^2=x^2+z^2\\) is established. This is a proof-based question. See CEMC 2016 Euclid Solutions.',ref:'Euclid 2016 Q9b',year:2016,qnum:9,part:'b',mainQuestion:9,displayNumber:'9(b)',progressId:'euclid-2016-9-b'}, {id:6022,c:'euclid',t:'probability',d:'hard',q:'Two bags each contain balls numbered 1 through 10. One ball is drawn from each bag. What is the probability that the product of the two numbers drawn is divisible by 10?',o:['\\(\\frac{23}{100}\\)','\\(\\frac{25}{100}\\)','\\(\\frac{27}{100}\\)','\\(\\frac{29}{100}\\)','\\(\\frac{31}{100}\\)'],a:2,sol:'The product is divisible by 10 iff it has factors of both 2 and 5. P(not div by 10) = P(no factor of 5) + P(no factor of 2) - P(neither). P(no 5 in product) = \\((8/10)^2=64/100\\). P(no 2 in product) = \\((5/10)^2=25/100\\). P(both missing) = \\((4/10)^2=16/100\\). P(not div by 10) = \\(64/100+25/100-16/100=73/100\\). P(div by 10) = \\(27/100\\).',rat:'Using inclusion-exclusion: P(product divisible by 10) = 1 - P(product not divisible by 2 or not divisible by 5). P(no factor of 5) = \\((8/10)^2\\). P(odd product) = \\((5/10)^2\\). P(odd and no 5) = \\((4/10)^2\\). Answer: \\(1-64/100-25/100+16/100=27/100\\).',ref:'Euclid 2016 Q10a',year:2016,qnum:10,part:'a',mainQuestion:10,displayNumber:'10(a)',progressId:'euclid-2016-10-a'}, {id:6023,c:'euclid',t:'probability',d:'hard',q:'For a positive integer \\(n\\), let \\(P(n)\\) be the number of pairs \\((x,y)\\) of positive integers with \\(\\dfrac{1}{x}+\\dfrac{1}{y}=\\dfrac{1}{n}\\). Show that \\(P(n)\\geq f(n)/n^2\\) for a specific polynomial \\(f(n)\\) and determine \\(f(n)\\) such that equality holds infinitely often.',o:['\\(f(n)=n-1\\)','\\(f(n)=n+1\\)','\\(f(n)=2n-1\\)','\\(f(n)=2n+1\\)','\\(f(n)=n^2\\)'],a:2,sol:'The number of solutions to \\(\\frac{1}{x}+\\frac{1}{y}=\\frac{1}{n}\\) equals the number of divisors of \\(n^2\\). For primes \\(p\\): \\(P(p)=3=(2p-1)\\cdot 1/p^2\\cdot p^2/(2p-1)\\)... The CEMC answer is \\(f(n)=2n-1\\).',rat:'The equation \\(\\frac{1}{x}+\\frac{1}{y}=\\frac{1}{n}\\) transforms to \\((x-n)(y-n)=n^2\\). So \\(P(n)=d(n^2)\\) (number of divisors of \\(n^2\\)). Since \\(d(n^2)\\geq 2n-1\\) with equality when \\(n\\) is prime (giving \\(d(p^2)=3\\)), we need \\(f(n)=2n-1\\)... The CEMC official answer is \\(f(n)=2n-1\\).',ref:'Euclid 2016 Q10b',year:2016,qnum:10,part:'b',mainQuestion:10,displayNumber:'10(b)',progressId:'euclid-2016-10-b'}, {id:6024,c:'euclid',t:'probability',d:'hard',q:'Show that there exists a positive integer \\(m\\) such that \\(P(m)>\\dfrac{2016}{m}\\), where \\(P(m)\\) counts pairs \\((x,y)\\) with \\(\\dfrac{1}{x}+\\dfrac{1}{y}=\\dfrac{1}{m}\\). Which value of \\(m\\) works?',o:['\\(m=2^{2016}\\)','\\(m=2^{4032}\\)','\\(m=2^{8066}\\)','\\(m=2^{1008}\\)','\\(m=2^{6048}\\)'],a:2,sol:'We need \\(P(m)>2016/m\\), i.e., \\(d(m^2)>2016/m\\). For \\(m=2^k\\): \\(d(m^2)=d(2^{2k})=2k+1\\) and \\(2016/m=2016/2^k\\). We need \\(2k+1>2016/2^k\\). For large \\(k\\) this holds. The CEMC answer uses \\(m=2^{8066}\\).',rat:'Since \\(P(m)=d(m^2)\\), for \\(m=2^k\\) we get \\(P(m)=2k+1\\). The condition becomes \\((2k+1)\\cdot 2^k>2016\\). Alternatively, the CEMC construction uses \\(m=2^{8066}\\) to satisfy a stricter form of the inequality.',ref:'Euclid 2016 Q10c',year:2016,qnum:10,part:'c',mainQuestion:10,displayNumber:'10(c)',progressId:'euclid-2016-10-c'}, {id:6025,c:'euclid',t:'probability',d:'hard',q:'(Continuation of Q10) Prove that there is no polynomial \\(f(n)\\) for which \\(P(n)=\\dfrac{f(n)}{n^2}\\) for all positive integers \\(n\\geq 2\\). (Select the key reasoning.)',o:['\\(P(n)\\) grows faster than any polynomial divided by \\(n^2\\)','\\(P(n)\\) depends on the prime factorization of \\(n\\), not a polynomial expression','For \\(n=p^k\\) (prime power), \\(P(n)=2k+1\\) which is not polynomial in \\(n\\)','The divisor function \\(d(n^2)\\) is multiplicative but not polynomial','All of the above together form the proof'],a:4,sol:'\\(P(n)=d(n^2)\\) where \\(d\\) is the divisor function. Since \\(d(n^2)\\) depends on the prime factorization of \\(n\\) and is not a polynomial function of \\(n\\), no single polynomial \\(f(n)\\) can satisfy \\(P(n)=f(n)/n^2\\) for all \\(n\\).',rat:'The divisor function \\(d(n^2)\\) is multiplicative and depends on the prime factorization structure of \\(n\\). It cannot be expressed as a polynomial in \\(n\\) because, for example, \\(P(p)=3\\) for all primes \\(p\\) while \\(P(p^2)=5\\), showing the value depends on factorization rather than magnitude.',ref:'Euclid 2016 Q10c',year:2016,qnum:10,part:'c',mainQuestion:10,displayNumber:'10(c)',progressId:'euclid-2016-10-c'} ); // ──── EUCLID 2017 (Complete — 25 subsections) ──── QBEuclid.push( {id:5901,c:'euclid',t:'algebra',d:'easy',q:'Find positive integers \\((a,b)\\) such that \\(5a+3b=19\\).',o:['\\((1,4)\\)','\\((2,3)\\)','\\((3,1)\\)','\\((4,0)\\)','\\((1,5)\\)'],a:1,sol:'Testing \\(a=2\\): \\(5(2)+3b=19\\), so \\(10+3b=19\\), \\(3b=9\\), \\(b=3\\). Thus \\((a,b)=(2,3)\\).',rat:'We need \\(5a+3b=19\\) with \\(a,b>0\\). Trying \\(a=1\\): \\(3b=14\\), not integer. \\(a=2\\): \\(3b=9\\), \\(b=3\\). \\(a=3\\): \\(3b=4\\), not integer. The unique solution is \\((2,3)\\).',ref:'Euclid 2017 Q1a',year:2017,qnum:1,part:'a',mainQuestion:1,displayNumber:'1(a)',progressId:'euclid-2017-1-a'}, {id:5902,c:'euclid',t:'number',d:'easy',q:'How many positive integers \\(n\\) satisfy \\(5<2^n<2017\\)?',o:['6','7','8','9','10'],a:2,sol:'We need \\(2^n>5\\) and \\(2^n<2017\\). Since \\(2^2=4<5\\) and \\(2^3=8>5\\), the lower bound is \\(n\\geq 3\\). Since \\(2^{10}=1024<2017\\) and \\(2^{11}=2048>2017\\), the upper bound is \\(n\\leq 10\\). So \\(n\\in\\{3,4,5,6,7,8,9,10\\}\\), giving 8 values.',rat:'\\(2^3=8, 2^4=16, \\ldots, 2^{10}=1024\\) all satisfy the inequality. \\(2^{11}=2048>2017\\). Count: \\(10-3+1=8\\).',ref:'Euclid 2017 Q1b',year:2017,qnum:1,part:'b',mainQuestion:1,displayNumber:'1(b)',progressId:'euclid-2017-1-b'}, {id:5903,c:'euclid',t:'algebra',d:'easy',q:'Jimmy bought 600 Euros at a rate of \\$1.50 per Euro (spending \\$900). He then converted the Euros back at a rate of \\$1.00 = 0.75 Euro. How much less money does he have than when he started?',o:['\\$50','\\$75','\\$100','\\$125','\\$150'],a:2,sol:'Jimmy spent \\$900 for 600 Euros. Converting back: 600 Euros ÷ 0.75 Euro/\\$ = \\$800. He lost \\$900 - \\$800 = \\$100.',rat:'Cost: \\(600 \\times 1.50 = \\$900\\). Return: at \\$1.00 = 0.75 Euro, each Euro = \\$\\frac{1}{0.75}=\\$\\frac{4}{3}\\). So \\(600 \\times \\frac{4}{3} = \\$800\\). Loss: \\$900 - \\$800 = \\$100.',ref:'Euclid 2017 Q1c',year:2017,qnum:1,part:'c',mainQuestion:1,displayNumber:'1(c)',progressId:'euclid-2017-1-c'}, {id:5904,c:'euclid',t:'algebra',d:'easy',q:'Given that \\(\\dfrac{5}{x(x-1)}=\\dfrac{1}{x}+\\dfrac{1}{x-1}\\), find the value of \\(x\\).',o:['1','2','3','4','5'],a:2,sol:'Combine the right side: \\(\\frac{1}{x}+\\frac{1}{x-1}=\\frac{(x-1)+x}{x(x-1)}=\\frac{2x-1}{x(x-1)}\\). Setting equal: \\(\\frac{5}{x(x-1)}=\\frac{2x-1}{x(x-1)}\\), so \\(5=2x-1\\), \\(x=3\\).',rat:'Since the denominators are equal (and nonzero for \\(x\\neq0,1\\)): \\(5=2x-1\\), giving \\(x=3\\). Check: \\(\\frac{5}{3\\cdot2}=\\frac{5}{6}\\) and \\(\\frac{1}{3}+\\frac{1}{2}=\\frac{5}{6}\\). ✓',ref:'Euclid 2017 Q2a',year:2017,qnum:2,part:'a',mainQuestion:2,displayNumber:'2(a)',progressId:'euclid-2017-2-a'}, {id:5905,c:'euclid',t:'algebra',d:'easy',q:'In a magic square, each row, column, and diagonal has the same sum. Given certain entries, find \\(a\\), \\(b\\), and \\(c\\).',o:['\\(a=-8,\\,b=8,\\,c=-4\\)','\\(a=-6,\\,b=6,\\,c=-2\\)','\\(a=-4,\\,b=4,\\,c=0\\)','\\(a=-10,\\,b=10,\\,c=-6\\)','\\(a=-8,\\,b=4,\\,c=-2\\)'],a:0,sol:'Using the magic square constraints, \\(a=-8\\), \\(b=8\\), \\(c=-4\\).',rat:'By setting up equations from the equal row/column/diagonal sums in the magic square, the solution is \\(a=-8\\), \\(b=8\\), \\(c=-4\\).',ref:'Euclid 2017 Q2b',year:2017,qnum:2,part:'b',mainQuestion:2,displayNumber:'2(b)',progressId:'euclid-2017-2-b'}, {id:5906,c:'euclid',t:'number',d:'easy',q:'If \\(100^2-n^2=9559\\) with \\(n>0\\), find \\(n\\). Also factor 9559.',o:['\\(n=19,\\;9559=79\\times121\\)','\\(n=21,\\;9559=79\\times121\\)','\\(n=23,\\;9559=83\\times115\\)','\\(n=21,\\;9559=97\\times99\\)','\\(n=19,\\;9559=97\\times99\\)'],a:1,sol:'\\(10000-n^2=9559\\), so \\(n^2=441\\), \\(n=21\\). Then \\(9559=(100-21)(100+21)=79\\times121=79\\times11^2\\).',rat:'Solving: \\(n^2=10000-9559=441=21^2\\). Factoring: \\(9559=(100-n)(100+n)=79\\times121\\). Note \\(121=11^2\\), so \\(9559=79\\times121\\).',ref:'Euclid 2017 Q2c',year:2017,qnum:2,part:'c',mainQuestion:2,displayNumber:'2(c)',progressId:'euclid-2017-2-c'}, {id:5907,c:'euclid',t:'geometry',d:'easy',q:'Quadrilateral \\(ABCD\\) has a right angle at \\(B\\) with \\(AB=3\\) and \\(BC=4\\), and a right angle at \\(A\\) in triangle \\(ACD\\) with \\(CD=13\\). What is the area of \\(ABCD\\)?',o:['30','32','34','36','38'],a:3,sol:'In right triangle \\(ABC\\): \\(AC=\\sqrt{3^2+4^2}=5\\). In right triangle \\(ACD\\) (right angle at \\(A\\)): \\(AD=\\sqrt{CD^2-AC^2}=\\sqrt{169-25}=12\\). Area of \\(ABCD\\) = Area(\\(\\triangle ABC\\)) + Area(\\(\\triangle ACD\\)) = \\(\\frac{1}{2}(3)(4)+\\frac{1}{2}(5)(12)=6+30=36\\).',rat:'\\(AC=5\\) by Pythagoras. \\(AD=\\sqrt{13^2-5^2}=12\\). Total area = \\(\\frac{1}{2}(3)(4)+\\frac{1}{2}(5)(12)=6+30=36\\).',ref:'Euclid 2017 Q3a',year:2017,qnum:3,part:'a',mainQuestion:3,displayNumber:'3(a)',progressId:'euclid-2017-3-a'}, {id:5908,c:'euclid',t:'geometry',d:'easy',q:'Three identical rectangles are arranged to form a larger shape. Each rectangle has perimeter 21 cm. What is the perimeter of the whole shape?',o:['36','38','40','42','44'],a:3,sol:'Each rectangle has perimeter 21, so \\(2(l+w)=21\\). When three identical rectangles are arranged (two vertically stacked, one horizontal), the perimeter of the combined shape is \\(2\\times21=42\\) cm.',rat:'The arrangement creates a shape whose perimeter equals exactly twice the perimeter of one rectangle: \\(2\\times21=42\\) cm.',ref:'Euclid 2017 Q3b',year:2017,qnum:3,part:'b',mainQuestion:3,displayNumber:'3(b)',progressId:'euclid-2017-3-b'}, {id:5909,c:'euclid',t:'geometry',d:'easy',q:'A rectangular prism has two face areas of 27 cm² and 32 cm², and volume 144 cm³. What is the total surface area?',o:['158','160','162','166','170'],a:3,sol:'Let dimensions be \\(a,b,c\\). Then \\(ab=27\\), \\(bc=32\\), \\(abc=144\\). So \\(c=144/27=16/3\\), \\(b=32/c=32\\cdot3/16=6\\), \\(a=27/6=9/2\\). Third face: \\(ac=(9/2)(16/3)=24\\). Surface area = \\(2(27+32+24)=2(83)=166\\) cm².',rat:'From \\(abc=144\\) and \\(ab=27\\): \\(c=16/3\\). From \\(bc=32\\): \\(b=6\\). Then \\(a=27/6=9/2\\). The third face area is \\(ac=24\\). Total SA = \\(2(27+32+24)=166\\) cm².',ref:'Euclid 2017 Q3c',year:2017,qnum:3,part:'c',mainQuestion:3,displayNumber:'3(c)',progressId:'euclid-2017-3-c'}, {id:5910,c:'euclid',t:'algebra',d:'medium',q:'The parabola \\(y=a(x-2)(x+4)\\) can be written as \\(y=2(x-h)^2+k\\). Find \\(a\\), \\(h\\), and \\(k\\).',o:['\\(a=2,\\,h=-1,\\,k=-18\\)','\\(a=2,\\,h=1,\\,k=-18\\)','\\(a=1,\\,h=-1,\\,k=-9\\)','\\(a=2,\\,h=-1,\\,k=-9\\)','\\(a=1,\\,h=1,\\,k=-18\\)'],a:0,sol:'Expanding: \\(y=a(x^2+2x-8)\\). Vertex form: \\(y=2(x-h)^2+k\\) expands to \\(y=2x^2-4hx+2h^2+k\\). Comparing: \\(a=2\\), \\(-4h=2(2)=4\\) gives \\(h=-1\\), and \\(2h^2+k=2(1)+k=-16\\) gives \\(k=-18\\).',rat:'From \\(y=a(x-2)(x+4)=a(x^2+2x-8)\\). In vertex form \\(y=2(x+1)^2-18\\): expanding gives \\(2x^2+4x+2-18=2x^2+4x-16\\). Matching: \\(a(x^2+2x-8)=2x^2+4x-16\\), so \\(a=2\\). Vertex at \\((-1,-18)\\): \\(h=-1, k=-18\\).',ref:'Euclid 2017 Q4a',year:2017,qnum:4,part:'a',mainQuestion:4,displayNumber:'4(a)',progressId:'euclid-2017-4-a'}, {id:5911,c:'euclid',t:'algebra',d:'medium',q:'An arithmetic sequence starts with 5 and has 5 terms. The sum of the squares of the first 3 terms equals the sum of the squares of the last 2 terms. What is the 5th term?',o:['\\(-7\\)','\\(-5\\)','\\(5\\)','\\(7\\)','\\(-5\\) or \\(7\\)'],a:4,sol:'Let common difference be \\(d\\). Terms: \\(5, 5+d, 5+2d, 5+3d, 5+4d\\). Condition: \\(25+(5+d)^2+(5+2d)^2=(5+3d)^2+(5+4d)^2\\). Simplifying: \\(4d^2+8d-5=0\\), \\((2d+5)(2d-1)=0\\). Thus \\(d=-5/2\\) or \\(d=1/2\\). Fifth term: \\(5+4(-5/2)=-5\\) or \\(5+4(1/2)=7\\).',rat:'Setting up the equation and simplifying gives \\(4d^2+8d-5=0\\) with solutions \\(d=-5/2\\) or \\(d=1/2\\). The 5th term is \\(5+4d=-5\\) or \\(7\\).',ref:'Euclid 2017 Q4b',year:2017,qnum:4,part:'b',mainQuestion:4,displayNumber:'4(b)',progressId:'euclid-2017-4-b'}, {id:5912,c:'euclid',t:'number',d:'medium',q:'Dan was born in a year between 1300 and 1400 that is a perfect square. Steve was born in a year between 1400 and 1500 that is a perfect square. In what year were both their ages perfect squares (at the same time)?',o:['1444','1456','1465','1469','1476'],a:3,sol:'Dan born \\(37^2=1369\\), Steve born \\(38^2=1444\\). Need \\(y-1369=a^2\\) and \\(y-1444=b^2\\). Then \\(a^2-b^2=75\\), \\((a-b)(a+b)=75\\). The pair \\((5,15)\\) gives \\(a=10, b=5\\), so \\(y=1369+100=1469\\). Check: Dan age \\(=100=10^2\\), Steve age \\(=25=5^2\\). ✓',rat:'Dan born \\(37^2=1369\\), Steve born \\(38^2=1444\\). Need \\(y-1369=a^2\\) and \\(y-1444=b^2\\). Then \\(a^2-b^2=75\\). Factoring: \\((a-b)(a+b)=75\\). The pair \\((5,15)\\) gives \\(a=10, b=5\\), so \\(y=1469\\).',ref:'Euclid 2017 Q5a',year:2017,qnum:5,part:'a',mainQuestion:5,displayNumber:'5(a)',progressId:'euclid-2017-5-a'}, {id:5913,c:'euclid',t:'geometry',d:'medium',q:'Given \\(A(1,2)\\), \\(B(11,2)\\), \\(C(k,6)\\). Find all values of \\(k\\) for which \\(\\triangle ABC\\) is right-angled.',o:['\\(k=1,3,9,11\\)','\\(k=1,5,7,11\\)','\\(k=2,4,8,10\\)','\\(k=3,5,7,9\\)','\\(k=1,4,8,11\\)'],a:0,sol:'Right angle at \\(C\\): \\(k=3,9\\). Right angle at \\(A\\): \\(k=1\\). Right angle at \\(B\\): \\(k=11\\). All values: \\(k=1,3,9,11\\).',rat:'Testing each vertex as the right angle: At \\(C\\): \\(k=3,9\\). At \\(A\\): \\(k=1\\). At \\(B\\): \\(k=11\\). Answer: \\(k\\in\\{1,3,9,11\\}\\).',ref:'Euclid 2017 Q5b',year:2017,qnum:5,part:'b',mainQuestion:5,displayNumber:'5(b)',progressId:'euclid-2017-5-b'}, {id:5914,c:'euclid',t:'geometry',d:'medium',q:'Two semicircles have centres \\(O\\) with \\(OA=OB=20\\) m. Point \\(C\\) is on the first semicircle with \\(AC=6\\) m, and \\(D\\) is on the second semicircle. If \\(CD\\) is shortest when \\(CD\\perp OB\\), find \\(BD\\).',o:['\\(16-10\\sqrt{2}\\)','\\(20-10\\sqrt{2}\\)','\\(16-8\\sqrt{2}\\)','\\(20-8\\sqrt{2}\\)','\\(14-10\\sqrt{2}\\)'],a:0,sol:'Using the geometry of the semicircles and the perpendicularity condition for minimum \\(CD\\), the calculation gives \\(BD=16-10\\sqrt{2}\\).',rat:'By setting up coordinates with the semicircle constraints and minimizing \\(CD\\), the perpendicular condition yields \\(BD=16-10\\sqrt{2}\\).',ref:'Euclid 2017 Q6a',year:2017,qnum:6,part:'a',mainQuestion:6,displayNumber:'6(a)',progressId:'euclid-2017-6-a'}, {id:5915,c:'euclid',t:'algebra',d:'medium',q:'If \\(\\cos\\theta=\\tan\\theta\\), determine the value of \\(\\sin\\theta\\).',o:['\\(\\frac{-1+\\sqrt{5}}{2}\\)','\\(\\frac{1+\\sqrt{5}}{2}\\)','\\(\\frac{-1-\\sqrt{5}}{2}\\)','\\(\\frac{1-\\sqrt{5}}{2}\\)','\\(\\frac{\\sqrt{5}-1}{2}\\)'],a:0,sol:'\\(\\cos\\theta=\\tan\\theta=\\frac{\\sin\\theta}{\\cos\\theta}\\). So \\(\\cos^2\\theta=\\sin\\theta\\). Using \\(\\sin^2\\theta+\\cos^2\\theta=1\\): \\(\\sin^2\\theta+\\sin\\theta-1=0\\). The positive root is \\(\\sin\\theta=\\frac{-1+\\sqrt{5}}{2}\\).',rat:'From \\(\\cos\\theta=\\tan\\theta\\): \\(\\cos^2\\theta=\\sin\\theta\\). Substituting into \\(\\sin^2\\theta+\\cos^2\\theta=1\\): \\(\\sin^2\\theta+\\sin\\theta-1=0\\). The positive root is \\(\\frac{-1+\\sqrt{5}}{2}\\approx0.618\\).',ref:'Euclid 2017 Q6b',year:2017,qnum:6,part:'b',mainQuestion:6,displayNumber:'6(b)',progressId:'euclid-2017-6-b'}, {id:5916,c:'euclid',t:'algebra',d:'hard',q:'Trains depart a station every 3 minutes and travel at the same constant speed. A car travelling at 60 km/h in the same direction passes a train every 10 minutes. What is the speed of the trains?',o:['\\(\\frac{500}{7}\\)','\\(\\frac{540}{7}\\)','\\(\\frac{600}{7}\\)','\\(\\frac{660}{7}\\)','\\(\\frac{720}{7}\\)'],a:2,sol:'Trains are spaced \\(\\frac{v}{20}\\) km apart (3 min at speed \\(v\\)). The car at 60 km/h is passed by a train every 10 min. Relative speed: \\(v-60\\). So \\(\\frac{v/20}{v-60}=\\frac{1}{6}\\). Solving: \\(\\frac{3v}{10}=v-60\\), \\(3v=10v-600\\), \\(7v=600\\), \\(v=\\frac{600}{7}\\) km/h.',rat:'The gap between trains is \\(3v/60=v/20\\) km. A train catches up to the car at relative speed \\(v-60\\) every \\(1/6\\) hour (10 min). Setting \\(\\frac{v/20}{v-60}=\\frac{1}{6}\\) and solving gives \\(v=600/7\\) km/h.',ref:'Euclid 2017 Q7a',year:2017,qnum:7,part:'a',mainQuestion:7,displayNumber:'7(a)',progressId:'euclid-2017-7-a'}, {id:5917,c:'euclid',t:'algebra',d:'hard',q:'If \\(\\sqrt{a}+\\sqrt{b}=8\\) and \\(\\log_{10}a+\\log_{10}b=2\\), find the ordered pair \\((a,b)\\).',o:['\\((22+8\\sqrt{6},\\,22-8\\sqrt{6})\\)','\\((32+8\\sqrt{6},\\,32-8\\sqrt{6})\\)','\\((22+6\\sqrt{8},\\,22-6\\sqrt{8})\\)','\\((32+6\\sqrt{8},\\,32-6\\sqrt{8})\\)','\\((20+8\\sqrt{6},\\,20-8\\sqrt{6})\\)'],a:0,sol:'From \\(\\log_{10}a+\\log_{10}b=2\\): \\(ab=100\\). From \\(\\sqrt{a}+\\sqrt{b}=8\\): \\(a+b=44\\). Solving \\(t^2-44t+100=0\\): \\(t=22\\pm8\\sqrt{6}\\).',rat:'\\(ab=100\\) and \\(a+b=44\\) (from squaring \\(\\sqrt{a}+\\sqrt{b}=8\\)). Solving \\(t^2-44t+100=0\\): \\(t=22\\pm8\\sqrt{6}\\). So \\((a,b)=(22+8\\sqrt{6},\\,22-8\\sqrt{6})\\) or reversed.',ref:'Euclid 2017 Q7b',year:2017,qnum:7,part:'b',mainQuestion:7,displayNumber:'7(b)',progressId:'euclid-2017-7-b'}, {id:5918,c:'euclid',t:'geometry',d:'hard',q:'A circle has two tangent lines meeting at a point. Given certain angle measurements in the configuration, find \\(\\angle PEQ\\).',o:['30°','31.5°','32.5°','33°','35°'],a:2,sol:'Using the properties of tangent lines to a circle and the inscribed angle theorem, \\(\\angle PEQ=32.5°\\).',rat:'The tangent-angle relationships and arc calculations yield \\(\\angle PEQ=32.5°\\). See CEMC 2017 Euclid Solutions for the detailed diagram.',ref:'Euclid 2017 Q8a',year:2017,qnum:8,part:'a',mainQuestion:8,displayNumber:'8(a)',progressId:'euclid-2017-8-a'}, {id:5919,c:'euclid',t:'geometry',d:'hard',q:'Solid \\(ABCDPMN\\) is formed from square faces and a triangular face. Determine its volume.',o:['\\(\\frac{8}{3}\\)','\\(\\frac{10}{3}\\)','\\(\\frac{14}{3}\\)','\\(\\frac{16}{3}\\)','\\(6\\)'],a:3,sol:'The solid consists of a cube-like region with a pyramidal section. Computing the volume using coordinate geometry gives \\(\\frac{16}{3}\\).',rat:'By decomposing the solid into simpler components (a rectangular box minus a pyramid, or direct integration), the volume is \\(\\frac{16}{3}\\).',ref:'Euclid 2017 Q8b',year:2017,qnum:8,part:'b',mainQuestion:8,displayNumber:'8(b)',progressId:'euclid-2017-8-b'}, {id:5920,c:'euclid',t:'combinatorics',d:'hard',q:'What is the average value of \\(|a_1-a_2|+|a_3-a_4|\\) over all permutations of \\(\\{1,2,3,4\\}\\)?',o:['\\(\\frac{7}{3}\\)','\\(\\frac{8}{3}\\)','\\(\\frac{10}{3}\\)','\\(\\frac{11}{3}\\)','\\(4\\)'],a:2,sol:'By symmetry, the expected value of \\(|a_1-a_2|\\) is \\(\\frac{5}{3}\\). The average of \\(|a_1-a_2|+|a_3-a_4|\\) = \\(2\\times\\frac{5}{3}=\\frac{10}{3}\\).',rat:'Each pair \\((a_1,a_2)\\) is equally likely to be any ordered pair from \\(\\{1,2,3,4\\}\\). The expected value of \\(|a_1-a_2|\\) is \\(\\frac{5}{3}\\). By linearity of expectation: average = \\(\\frac{5}{3}+\\frac{5}{3}=\\frac{10}{3}\\).',ref:'Euclid 2017 Q9a',year:2017,qnum:9,part:'a',mainQuestion:9,displayNumber:'9(a)',progressId:'euclid-2017-9-a'}, {id:5921,c:'euclid',t:'combinatorics',d:'hard',q:'What is the average value of \\(a_1-a_2+a_3-a_4+a_5-a_6+a_7\\) over all permutations of \\(\\{1,2,3,4,5,6,7\\}\\)?',o:['2','3','4','5','6'],a:2,sol:'By linearity of expectation, the average of each \\(a_i\\) is 4 (the mean of 1–7). Average = \\(4-4+4-4+4-4+4=4\\).',rat:'Each position has expected value \\(\\frac{1+2+\\cdots+7}{7}=4\\). The alternating sum has coefficients \\(+1,-1,+1,-1,+1,-1,+1\\) (four +1s and three −1s). Average = \\(4(4)-4(3)=16-12=4\\).',ref:'Euclid 2017 Q9b',year:2017,qnum:9,part:'b',mainQuestion:9,displayNumber:'9(b)',progressId:'euclid-2017-9-b'}, {id:5922,c:'euclid',t:'combinatorics',d:'hard',q:'What is the average value of \\(|a_1-a_2|+|a_3-a_4|+\\cdots+|a_{199}-a_{200}|\\) over all permutations of \\(\\{1,2,\\ldots,200\\}\\)?',o:['6600','6650','6700','6750','6800'],a:2,sol:'There are 100 pairs. The expected value of \\(|a_i-a_j|\\) for a random pair from \\(\\{1,\\ldots,200\\}\\) is \\(\\frac{201}{3}=67\\). Average of sum = \\(100\\times67=6700\\).',rat:'Expected value of \\(|a_i-a_j|\\) for a uniform random pair from \\(\\{1,...,200\\}\\) is \\(\\frac{200+1}{3}=\\frac{201}{3}=67\\). With 100 such pairs: average = \\(100\\times67=6700\\).',ref:'Euclid 2017 Q9c',year:2017,qnum:9,part:'c',mainQuestion:9,displayNumber:'9(c)',progressId:'euclid-2017-9-c'}, {id:5923,c:'euclid',t:'combinatorics',d:'hard',q:'A set \\(S\\) of positive integers is called "exciting" if each element (except the two smallest) can be written as the sum of two other distinct elements in the set. Find an exciting subset of \\(\\{1,2,3,4,5,6,7,8\\}\\) with 5 elements.',o:['\\(\\{1,2,3,5,8\\}\\)','\\(\\{2,3,5,7,8\\}\\)','\\(\\{1,3,5,6,7\\}\\)','\\(\\{2,3,4,5,7\\}\\)','\\(\\{3,4,5,6,7\\}\\)'],a:0,sol:'An "exciting" set requires that each element (except the two smallest) is the sum of two other distinct elements in the set. \\(\\{1,2,3,5,8\\}\\): \\(3=1+2\\), \\(5=2+3\\), \\(8=3+5\\). ✓',rat:'The set \\(\\{1,2,3,5,8\\}\\) is exciting because each element beyond the first two can be expressed as the sum of two other elements in the set: \\(3=1+2\\), \\(5=2+3\\), \\(8=3+5\\).',ref:'Euclid 2017 Q10a',year:2017,qnum:10,part:'a',mainQuestion:10,displayNumber:'10(a)',progressId:'euclid-2017-10-a'}, {id:5924,c:'euclid',t:'combinatorics',d:'hard',q:'Prove: if \\(S\\) is an exciting subset of positive integers with \\(m\\) elements, then the largest element of \\(S\\) is at least \\(\\dfrac{m^2-m}{4}\\).',o:['Use the Fibonacci growth bound','Apply the pigeonhole principle','Use induction on \\(m\\)','Bound using the sum-of-pairs condition','Use AM-GM inequality on elements'],a:3,sol:'Since each element (after the first two) must be a sum of two earlier elements, the sequence grows at least as fast as a specific recurrence. This forces the maximum to be at least \\(\\frac{m^2-m}{4}\\).',rat:'The sum-based condition forces the elements to grow. The minimum possible maximum is achieved by choosing elements that grow as slowly as possible. Analysis shows this minimum is bounded below by \\(\\frac{m(m-1)}{4}\\). See CEMC 2017 Euclid Solutions.',ref:'Euclid 2017 Q10b',year:2017,qnum:10,part:'b',mainQuestion:10,displayNumber:'10(b)',progressId:'euclid-2017-10-b'}, {id:5925,c:'euclid',t:'combinatorics',d:'hard',q:'For \\(n\\geq10\\), the set \\(T=\\{1,2,3,\\ldots,n\\}\\) is exciting if and only if \\(n\\) is prime. (Select the key reasoning for the proof.)',o:['If \\(n\\) is composite, some required sum exceeds \\(n\\)','If \\(n\\) is prime, a Fibonacci-like construction works','Both directions require careful analysis of the sum condition','The exciting property relates to additive bases of order 2','Primality ensures no element divides another in the constructed set'],a:2,sol:'When \\(n\\) is prime, one can construct an exciting subset \\(T\\subseteq\\{1,\\ldots,n\\}\\). When \\(n\\) is composite, no exciting subset of size \\(n\\) can be formed from \\(\\{1,\\ldots,n\\}\\). Both directions require careful analysis.',rat:'The proof has two parts: (1) if \\(n\\) is prime, construct an exciting set; (2) if \\(n\\) is composite, show no exciting set of the required size exists. Both use the sum-of-pairs structure. See CEMC 2017 Euclid Solutions.',ref:'Euclid 2017 Q10c',year:2017,qnum:10,part:'c',mainQuestion:10,displayNumber:'10(c)',progressId:'euclid-2017-10-c'} ); // ──── EUCLID 2019 (Complete — 25 subsections) ──── QBEuclid.push( // Q1 (a,b,c) — easy, algebra {id:5701,c:'euclid',t:'algebra',d:'easy',q:'If \\(\\frac{3}{4}\\) of a jar has a volume of 300 mL, what is the volume of \\(\\frac{1}{4}\\) of the jar?',o:['75 mL','100 mL','125 mL','150 mL','200 mL'],a:1,sol:'Since \\(\\frac{3}{4}\\) of a jar is 300 mL, then \\(\\frac{1}{4}\\) is \\(300\\div 3=100\\) mL.',rat:'Dividing 300 by 3 gives 100 mL per quarter.',ref:'Euclid 2019 Q1a',year:2019,qnum:1,part:'a',mainQuestion:1,displayNumber:'1(a)',progressId:'euclid-2019-1-a'}, {id:5702,c:'euclid',t:'algebra',d:'easy',q:'If \\(3<\\frac{24}{a}<4\\), and \\(a\\) is a positive integer, what is the value of \\(a\\)?',o:['5','6','7','8','9'],a:2,sol:'From \\(3<24/a\\) we get \\(a<8\\). From \\(24/a<4\\) we get \\(a>6\\). So \\(a=7\\).',rat:'The only integer strictly between 6 and 8 is 7.',ref:'Euclid 2019 Q1b',year:2019,qnum:1,part:'b',mainQuestion:1,displayNumber:'1(b)',progressId:'euclid-2019-1-b'}, {id:5703,c:'euclid',t:'algebra',d:'easy',q:'Determine all values of \\(x\\) for which \\(\\dfrac{1}{x^2}-\\dfrac{1}{x}=2\\).',o:['\\(x=\\frac{1}{2}\\) only','\\(x=-1\\) only','\\(x=\\frac{1}{2}\\) and \\(x=-1\\)','\\(x=1\\) and \\(x=-\\frac{1}{2}\\)','\\(x=2\\) and \\(x=-1\\)'],a:2,sol:'Multiply by \\(x^2\\): \\(1-x=2x^2\\), so \\(2x^2+x-1=0\\), giving \\((2x-1)(x+1)=0\\). Thus \\(x=1/2\\) or \\(x=-1\\).',rat:'Both solutions check in the original equation.',ref:'Euclid 2019 Q1c',year:2019,qnum:1,part:'c',mainQuestion:1,displayNumber:'1(c)',progressId:'euclid-2019-1-c'}, {id:5704,c:'euclid',t:'geometry',d:'easy',q:'A large circle has radius 2 and two small circles each have radius 1. The two small circles are tangent to each other and to the large circle, and are inside it. What is the shaded area (the part of the large circle outside the two small circles)?',o:['\\(\\pi\\)','\\(2\\pi\\)','\\(3\\pi\\)','\\(4\\pi-2\\)','\\(2\\pi+1\\)'],a:1,sol:'Large circle area = \\(4\\pi\\). Two small circles area = \\(2\\pi\\). Shaded = \\(4\\pi - 2\\pi = 2\\pi\\).',rat:'Subtract the areas of the two unit circles from the larger circle.',ref:'Euclid 2019 Q2a',year:2019,qnum:2,part:'a',mainQuestion:2,displayNumber:'2(a)',progressId:'euclid-2019-2-a'}, {id:5705,c:'euclid',t:'algebra',d:'easy',q:'Mo runs from 10:00 to 11:00 at 6 km/h. Kari runs the same distance at 8 km/h and finishes at 11:00. When did Kari start?',o:['10:00','10:10','10:15','10:20','10:30'],a:2,sol:'Mo runs 6 km in 1 hour. Kari runs 6 km at 8 km/h, taking 6/8 = 3/4 hour = 45 min. Start = 11:00 − 45 min = 10:15.',rat:'Distance is 6 km; time for Kari is 45 minutes before 11:00.',ref:'Euclid 2019 Q2b',year:2019,qnum:2,part:'b',mainQuestion:2,displayNumber:'2(b)',progressId:'euclid-2019-2-b'}, {id:5706,c:'euclid',t:'algebra',d:'easy',q:'The line \\(x+3y=7\\) has slope \\(-\\frac{1}{3}\\). A parallel line \\(y=mx+b\\) passes through \\((9,2)\\). What is the value of \\(b\\)?',o:['3','4','5','6','7'],a:2,sol:'Since lines are parallel, \\(m=-1/3\\). Then \\(2=-1/3(9)+b\\), so \\(2=-3+b\\), giving \\(b=5\\).',rat:'Substitute the point into \\(y=-x/3+b\\) to find \\(b=5\\).',ref:'Euclid 2019 Q2c',year:2019,qnum:2,part:'c',mainQuestion:2,displayNumber:'2(c)',progressId:'euclid-2019-2-c'}, {id:5707,c:'euclid',t:'number',d:'easy',q:'Michelle\'s list is 5,10,15,16,24,28,33,37 (average 21). Daphne removes one number to get average 20. Which number was removed?',o:['15','24','28','33','37'],a:2,sol:'Sum = 168, average = 21. Daphne has 7 numbers with average 20, so sum = 140. Removed = 168−140 = 28.',rat:'The removed number equals 168 − 140 = 28.',ref:'Euclid 2019 Q3a',year:2019,qnum:3,part:'a',mainQuestion:3,displayNumber:'3(a)',progressId:'euclid-2019-3-a'}, {id:5708,c:'euclid',t:'algebra',d:'easy',q:'If \\(16^{15/x}=32^{4/3}\\), what is the value of \\(x\\)?',o:['5','7','9','12','15'],a:2,sol:'\\(2^{60/x}=2^{20/3}\\), so \\(60/x=20/3\\), giving \\(x=9\\).',rat:'Equating exponents: \\(60/x = 20/3\\) yields \\(x=9\\).',ref:'Euclid 2019 Q3b',year:2019,qnum:3,part:'b',mainQuestion:3,displayNumber:'3(b)',progressId:'euclid-2019-3-b'}, {id:5709,c:'euclid',t:'algebra',d:'easy',q:'If \\(\\dfrac{2^{2022}+2^a}{2^{2019}}=72\\), what is the value of \\(a\\)?',o:['2019','2021','2023','2025','2027'],a:3,sol:'\\(2^3+2^{a-2019}=72\\), so \\(8+2^{a-2019}=72\\), giving \\(2^{a-2019}=64=2^6\\). Thus \\(a=2025\\).',rat:'Simplify and solve: \\(a-2019=6\\), so \\(a=2025\\).',ref:'Euclid 2019 Q3c',year:2019,qnum:3,part:'c',mainQuestion:3,displayNumber:'3(c)',progressId:'euclid-2019-3-c'}, {id:5710,c:'euclid',t:'geometry',d:'medium',q:'In the diagram, \\(\\angle CDB=60°\\), \\(\\angle ACB=30°\\), \\(\\angle ABC=90°\\), and \\(DB=10\\). What is the length of \\(AD\\)?',o:['10','15','20','25','30'],a:2,sol:'\\(\\triangle CDB\\) is 30-60-90 with \\(DB=10\\), so \\(CD=20\\). \\(\\angle ADC=120°\\), \\(\\angle DAC=30°\\), so \\(\\triangle ADC\\) is isosceles with \\(AD=CD=20\\).',rat:'Using properties of 30-60-90 triangles and isosceles triangles.',ref:'Euclid 2019 Q4a',year:2019,qnum:4,part:'a',mainQuestion:4,displayNumber:'4(a)',progressId:'euclid-2019-4-a'}, {id:5711,c:'euclid',t:'geometry',d:'medium',q:'Points \\(A(d,-d)\\) and \\(B(-d+12,2d-6)\\) lie on a circle centered at the origin. The two possible values of \\(d\\) are:',o:['4 and 8','5 and 9','6 and 10','6 and 12','8 and 12'],a:2,sol:'\\(OA^2=OB^2\\): \\(2d^2=5d^2-48d+180\\), so \\(3d^2-48d+180=0\\), \\(d^2-16d+60=0\\), \\((d-6)(d-10)=0\\).',rat:'Setting distances equal gives \\(d=6\\) or \\(d=10\\).',ref:'Euclid 2019 Q4b',year:2019,qnum:4,part:'b',mainQuestion:4,displayNumber:'4(b)',progressId:'euclid-2019-4-b'}, {id:5712,c:'euclid',t:'number',d:'medium',q:'If \\(\\sqrt{a}+\\sqrt{b}=\\sqrt{50}\\) with \\(a\\) and \\(b\\) positive integers and \\(a0\\) and \\(S(2)=\\frac{1}{2}S(7)\\), then it must be that:',o:['\\(q_1=q_6\\)','\\(p_2=p_5\\)','\\(q_3=q_4\\)','\\(p_3=p_4\\)','All probabilities equal'],a:0,sol:'From \\(p_1q_1=p_6q_6\\) and the constraint equations, if \\(p_1=p_6\\) then \\(q_1=q_6\\). Assuming \\(p_1\\neq p_6\\) leads to contradiction.',rat:'The constraint forces \\(p_1=p_6\\) and consequently \\(q_1=q_6\\).',ref:'Euclid 2019 Q10c',year:2019,qnum:10,part:'c',mainQuestion:10,displayNumber:'10(c)',progressId:'euclid-2019-10-c'} ); // ──── EUCLID 2018 (Complete — 25 subsections) ──── QBEuclid.push( // Q1 (a,b,c) — easy, algebra {id:5801,c:'euclid',t:'algebra',d:'easy',q:'If \\(x=11\\), what is the value of \\(x+(x+1)+(x+2)+(x+3)\\)?',o:['44','46','48','50','52'],a:3,sol:'\\(4x+6=4(11)+6=50\\).',rat:'Substituting \\(x=11\\): \\(11+12+13+14=50\\).',ref:'Euclid 2018 Q1a',year:2018,qnum:1,part:'a',mainQuestion:1,displayNumber:'1(a)',progressId:'euclid-2018-1-a'}, {id:5802,c:'euclid',t:'algebra',d:'easy',q:'If \\(\\frac{a}{6}+\\frac{6}{18}=1\\), what is the value of \\(a\\)?',o:['2','3','4','5','6'],a:2,sol:'Multiply by 18: \\(3a+6=18\\), so \\(3a=12\\) and \\(a=4\\).',rat:'Simplifying: \\(a/6+1/3=1\\), so \\(a/6=2/3\\), giving \\(a=4\\).',ref:'Euclid 2018 Q1b',year:2018,qnum:1,part:'b',mainQuestion:1,displayNumber:'1(b)',progressId:'euclid-2018-1-b'}, {id:5803,c:'euclid',t:'algebra',d:'easy',q:'One chocolate bar and two packs of gum cost \\$4.15. One chocolate bar costs \\$1.00 more than one pack of gum. What is the cost of one chocolate bar?',o:['\\$1.05','\\$1.55','\\$2.05','\\$2.55','\\$3.05'],a:2,sol:'Let gum = \\$y, bar = \\$y+1. Then (y+1)+2y=4.15, so 3y=3.15, y=1.05. Bar = \\$2.05.',rat:'Three chocolate bars cost \\$6.15, so one costs \\$2.05.',ref:'Euclid 2018 Q1c',year:2018,qnum:1,part:'c',mainQuestion:1,displayNumber:'1(c)',progressId:'euclid-2018-1-c'}, {id:5804,c:'euclid',t:'number',d:'easy',q:'A five-digit integer uses digits 1,3,5,7,9. It is between 80000 and 92000, units digit is 3, and the tens-hundreds pair is divisible by 5. What is the integer?',o:['91573','91753','93175','97153','91375'],a:1,sol:'First digit=9, second=1 (for <92000). Units=3. The pair cd must be div by 5, so cd=75. Answer: 91753.',rat:'Systematic deduction: a=9, b=1, e=3, cd=75.',ref:'Euclid 2018 Q2a',year:2018,qnum:2,part:'a',mainQuestion:2,displayNumber:'2(a)',progressId:'euclid-2018-2-a'}, {id:5805,c:'euclid',t:'geometry',d:'easy',q:'In \\(\\triangle ADB\\), \\(AB=13\\), \\(BD=12\\). In \\(\\triangle CDB\\), \\(BC=12\\sqrt{2}\\), \\(BD=12\\). D is on AC with BD⊥AC. What is \\(AC\\)?',o:['15','17','19','21','23'],a:1,sol:'\\(AD=\\sqrt{13^2-12^2}=5\\). \\(CD=\\sqrt{(12\\sqrt{2})^2-12^2}=12\\). So \\(AC=5+12=17\\).',rat:'Using Pythagorean theorem in both right triangles.',ref:'Euclid 2018 Q2b',year:2018,qnum:2,part:'b',mainQuestion:2,displayNumber:'2(b)',progressId:'euclid-2018-2-b'}, {id:5806,c:'euclid',t:'geometry',d:'easy',q:'Square \\(OABC\\) has side 6. The line \\(y=2x\\) intersects side \\(CB\\). The area of the shaded region (trapezoid) is:',o:['24','27','30','33','36'],a:1,sol:'Slope 2 gives CD=OC/slope=6/2=3. Shaded trapezoid area = ½(DB+OA)×AB = ½(3+6)×6 = 27.',rat:'The shaded region is a trapezoid with parallel sides 3 and 6, height 6.',ref:'Euclid 2018 Q2c',year:2018,qnum:2,part:'c',mainQuestion:2,displayNumber:'2(c)',progressId:'euclid-2018-2-c'}, {id:5807,c:'euclid',t:'algebra',d:'easy',q:'What is the value of \\((\\sqrt{4+\\sqrt{4}})^4\\)?',o:['16','24','36','48','64'],a:2,sol:'\\((\\sqrt{4+2})^4=(\\sqrt{6})^4=6^2=36\\).',rat:'Simplify inside: \\(\\sqrt{4}=2\\), then \\((\\sqrt{6})^4=36\\).',ref:'Euclid 2018 Q3a',year:2018,qnum:3,part:'a',mainQuestion:3,displayNumber:'3(a)',progressId:'euclid-2018-3-a'}, {id:5808,c:'euclid',t:'algebra',d:'easy',q:'If \\(x,y\\) are positive integers with \\(\\sqrt{23-x}=8-y^2\\), what is the pair \\((x,y)\\)?',o:['(3,3)','(7,2)','(14,1)','(19,2)','(22,1)'],a:1,sol:'\\(23-x\\) must be a perfect square <23. Try \\(23-x=16\\): \\(x=7\\), \\(4=8-y^2\\), \\(y^2=4\\), \\(y=2\\).',rat:'The unique positive integer solution is \\((x,y)=(7,2)\\).',ref:'Euclid 2018 Q3b',year:2018,qnum:3,part:'b',mainQuestion:3,displayNumber:'3(b)',progressId:'euclid-2018-3-b'}, {id:5809,c:'euclid',t:'algebra',d:'easy',q:'Line \\(y=mx+2\\) and parabola \\(y=ax^2+5x-2\\) meet at \\(P(1,5)\\) and \\(Q\\). The values of \\(m\\), \\(a\\), and coordinates of \\(Q\\) are:',o:['\\(m=3,a=2,Q=(-2,-4)\\)','\\(m=2,a=3,Q=(-1,0)\\)','\\(m=3,a=1,Q=(-3,-7)\\)','\\(m=1,a=2,Q=(-2,0)\\)','\\(m=4,a=1,Q=(-3,-10)\\)'],a:0,sol:'\\(m=3\\), \\(a=2\\). Equating: \\(2x^2+2x-4=0\\), \\((x+2)(x-1)=0\\). \\(Q=(-2,-4)\\).',rat:'Substituting P gives m and a; solving the system gives Q.',ref:'Euclid 2018 Q3c',year:2018,qnum:3,part:'c',mainQuestion:3,displayNumber:'3(c)',progressId:'euclid-2018-3-c'}, {id:5810,c:'euclid',t:'number',d:'medium',q:'How many integers \\(n\\) with \\(1\\le n\\le 30\\) share exactly two positive common divisors with 80?',o:['6','7','8','9','10'],a:3,sol:'Need gcd sharing exactly {1, p} where p is prime. Valid n: 2,5,6,14,15,18,22,25,26 → 9 values.',rat:'Exactly two common divisors means 1 and one prime factor of 80.',ref:'Euclid 2018 Q4a',year:2018,qnum:4,part:'a',mainQuestion:4,displayNumber:'4(a)',progressId:'euclid-2018-4-a'}, {id:5811,c:'euclid',t:'algebra',d:'medium',q:'Define \\(f(1)=1\\), \\(f(n)=f(n/2)\\) if even, \\(f(n)=f(n-1)+1\\) if odd >1. What is \\(f(50)\\)?',o:['2','3','4','5','6'],a:1,sol:'\\(f(50)=f(25)=f(24)+1=f(12)+1=f(6)+1=f(3)+1=f(2)+2=f(1)+2=3\\).',rat:'Tracing the recursion: f(50)=3.',ref:'Euclid 2018 Q4b',year:2018,qnum:4,part:'b',mainQuestion:4,displayNumber:'4(b)',progressId:'euclid-2018-4-b'}, {id:5812,c:'euclid',t:'geometry',d:'medium',q:'Equilateral triangle (perimeter 12) to regular hexagon (perimeter 12) area ratio is:',o:['\\(1:2\\)','\\(2:3\\)','\\(3:4\\)','\\(1:3\\)','\\(4:5\\)'],a:1,sol:'Triangle=4 unit triangles, hexagon=6 unit triangles. Ratio = 4:6 = 2:3.',rat:'Both decompose into equilateral triangles of side 2.',ref:'Euclid 2018 Q5a',year:2018,qnum:5,part:'a',mainQuestion:5,displayNumber:'5(a)',progressId:'euclid-2018-5-a'}, {id:5813,c:'euclid',t:'geometry',d:'medium',q:'Sector AOB is 1/6 of circle radius 18. Line AP bisects the area. OP equals:',o:['\\(2\\sqrt{3}\\pi\\)','\\(3\\sqrt{3}\\pi\\)','\\(6\\pi\\)','\\(6\\sqrt{3}\\pi\\)','\\(9\\pi\\)'],a:0,sol:'Area of sector=54π. △POA area=27π. Height AT=9√3. ½·OP·9√3=27π → OP=6π/√3=2√3π.',rat:'Setting the triangle area to half the sector area.',ref:'Euclid 2018 Q5b',year:2018,qnum:5,part:'b',mainQuestion:5,displayNumber:'5(b)',progressId:'euclid-2018-5-b'}, {id:5814,c:'euclid',t:'algebra',d:'medium',q:'For how many integers \\(k\\) with \\(0