HMMT February

Let $n=\underline{a} \underline{b} \underline{b} \underline{a}$. If $a, b \leq 4$ then there are no carries in the multiplication $n \times 2$, and $2n=(2a)(2b)(2b)(2a)$ is a palindrome. We shall show conversely that if $n$ and $2n$ are palindromes, then necessarily $a, b \leq 4$. Hence the answer to the problem is $4 \times 5=\mathbf{20}$ (because $a$ cannot be zero). If $a \geq 5$ then $2n$ is a five-digit number whose most significant digit is 1, but because $2n$ is even, its least significant digit is even, contradicting the assumption that $2n$ is a palindrome. Therefore $a \leq 4$. Consequently $2n$ is a four-digit number, and its tens and hundreds digits must be equal. Because $a \leq 4$, there is no carry out of the ones place in the multiplication $n \times 2$, and therefore the tens digit of $2n$ is the ones digit of $2b$. In particular, the tens digit of $2n$ is even. But if $b \geq 5$, the carry out of the tens place makes the hundreds digit of $2n$ odd, which is impossible. Hence $b \leq 4$ as well.