A palindromic number reads the same both ways. The largest palindrome made from the product of two 2-digit numbers is 9009 = 91 × 99.
Find the largest palindrome made from the product of two 3-digit numbers.

"Palindromeness" is very different from most interesting properties of numbers; it is an artifact of one of many possible *representations* of the number. In this case, the representation in base 10.
Let's start by nailing down a definition for "palindrome".
Suppose $A = \sum_{k=0}^{t-1} a_k 10^k$ is a $t$-digit number base 10 (that is, the $a_k$ are integers between 0 and 9 (inclusive) and $a_{t-1}$ is not zero). We say $A$ is a *palindrome* if $a_{t-1-k} = a_k$ for all $0 \leq k < t$.

Let $t,u,v \geq 1$ be natural numbers, and let $A = \sum_{k=0}^{t-1} a_k 10^k$ and $B = \sum_{k=0}^{u-1} b_k 10^k$ be $t$- and $u$-digit palindromes, respectively. That is, $a_k = a_{t-1-k}$ when $0 \leq k < t$ and $b_k = b_{u-1-k}$ when $0 \leq k < u$. Then $$M = A + 10^{t+v} B + 10^{t+u+2v} A$$ is a $(2t+u+2v)$-digit palindrome.

Note that
$$\begin{eqnarray*}
M
& = & A + 10^{t+v} B + 10^{2t+u+2v} A \\
& = & \sum_{k=0}^{t-1} a_k 10^k + 10^{t+v} \sum_{k=0}^{u-1} b_k 10^k + 10^{2t+u+2v} \sum_{k=0}^{t-1} a_k 10^k \\
& = & \sum_{k=0}^{t-1} a_k 10^k + \sum_{k=0}^{u-1} b_k 10^{t+v+k} + \sum_{k=0}^{t-1} a_k 10^{t+u+2v+k} \\
& = & \sum_{k=0}^{t-1} a_k 10^k + \sum_{k=t+v}^{t+u+v-1} b_{k-t-v} 10^k + \sum_{k=t+u+2v}^{2t+u+2v-1} a_{k-t-u-2v} 10^k \\
& = & \sum_{k=0}^{2t+u+2v-1} e_k 10^k
\end{eqnarray*}$$
where
$$e_k = \left\{ \begin{array}{ll} a_k & \mathrm{if}\ 0 \leq k < t \\ 0 & \mathrm{if}\ t \leq k < t+v \\ b_{k-t-v} & \mathrm{if}\ t+v \leq k < t+u+v \\ 0 & \mathrm{if}\ t+u+v \leq k < t+u+2v \\ a_{k-t-u-2v} & \mathrm{if}\ t+u+2v \leq k < 2t+u+2v. \end{array} \right.$$
Certainly $M$ has $2t+u+2v$ digits. To see that $M$ is a palindrome, we need to check that $$e_{2t+u+2v-1-k} = e_k$$ for $0 \leq k < 2t+u+2v$. We will break this interval into 5 subintervals.
1. Suppose $0 \leq k < t$. Then $0 \leq t-1-k < t$, and so $t+u+2v \leq 2t+u+2v-1-k < 2t+u+2v$. So $$e_{2t+u+2v-1-k} = a_{2t+u+2v-1-k-t-u-2v} = a_{t-1-k} = a_k = e_k.$$
2. Suppose $t \leq k < t+v$. Then $0 \leq t+v-1-k < v$, and so $t+u+v \leq 2t+u+2v-1-k < t+u+2v$. So $$e_{2t+u+2v-1-k} = 0 = e_k.$$
3. Suppose $t+v \leq k < t+u+v$. Then $0 \leq t+u+v-1-k < u$, and so $t+v \leq 2t+u+2v-1-k < t+u+v$. So $$\begin{eqnarray*} e_{2t+u+2v-1-k} & = & b_{2t+u+2v-1-k-t-v} \\ & = & b_{t+u+v-1-k} \\ & = & b_{u-1-t-u-v+1-k} \\ & = & b_{k-t-v} \\ & = & e_{k}. \end{eqnarray*}$$
4. Suppose $t+u+v \leq k < t+u+2v$. Then $0 \leq t+u+2v-1-k < v$, and so $t \leq 2t+u+2v-1-k < t+v$. So $$e_{2t+u+2v-1-k} = 0 = e_k.$$
5. Suppose $t+u+2v \leq k < 2t+u+2v$. Then $0 \leq 2t+u+2v-1-k < t$. So $$\begin{eqnarray*} e_{2t+u+2v-1-k} & = & a_{2t+u+2v-1-k} \\ & = & a_{t-1-2t-u-2v+i+k} \\ & = & a_{k+t-u-2v} \\ & = & e_k. \end{eqnarray*}$$
Thus $M$ is a $(2t+u+2v)$-digit palindrome.

Let $A$ and $B$ be $t$-digit numbers such that both $AB$ and $2AB$ are $2t$-digit palindromes. Let $v$ be a positive integer, and define $$H_v(A) = A(1 + 10^{2t+v})\ \mathrm{and}\ K_v(B) = B(1 + 10^{2t+v}).$$ Then $H_v$ and $K_v$ are $(3t+v)$-digit numbers and $H_vK_v$ is a $(6t+2v)$-digit palindrome.

Expanding $H_vK_v$, we have $$H_vK_v = AB + 10^{2t+v}(2AB) + 10^{2t+2t+2v}AB.$$ The previous theorem applies, and $H_vK_v$ is a $(6t+2v)$-digit palindrome.

Let $t$ and $m$ be positive natural numbers, and define $Q_{t,m} = \sum_{k=0}^{t-1} 10^{mk}$. Now define $$A_{t,m} = Q_{2t,m},$$ $$B_{t,m} = 1 + 10^{tm}(10^m-1)Q_{t,m},$$ and $$C_t = Q_{t,m}(1 + 10^{3tm}).$$
Then $A_t$ is a $(2tm - m + 1)$-digit number, and $B_t$ is a $2tm$-digit number, and both $A_tB_t$ and $2A_tB_t$ are $(4tm - m + 1)$-digit palindromes.

Note that $A_{t,m} = Q_{t,m}(1 + 10^{tm})$ and $$Q_{t,m} = \frac{10^{tm} - 1}{10^m-1}.$$ Expanding $A_{t,m}B_{t,m}$, then, we have $$A_{t,m}B_{t,m} = Q_{t,m}(1 + 10^{tm})(1 - 10^{tm} + 10^{2tm}) = C_{t,m}$$ as needed.
To see the digit counts, note that $Q_{t,m}$ has $(tm-m+1)$ digits. Note also that all the digits of $C_{t,m}$ are 1, so that $2C_{t,m}$ is also a palindrome.

Let $k \geq 2$. Then among the $k$-digit numbers, there exists a pair $A$ and $B$ such that $AB$ is a $2k$-digit palindrome.
In particular, the largest such palindrome has $2k$ digits.