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Brazil competition problems

Number theory Difficulty 5.6 AIME, harder Prove it Brazil

A natural number is a palindrome when one obtains the same number when writing its digits in reverse order. For example, $481184$ , $131$ and $2$ are palindromes.
Determine all pairs $(m, n)$ of positive integers such that $\overbrace{111\ldots1}^{m \text{ ones}} \times \overbrace{111\ldots1}^{n \text{ ones}}$ is a palindrome.

Solution

Note that $N = \overbrace{111\ldots1}^{m \text{ ones}} \times \overbrace{111\ldots1}^{n \text{ ones}}$ has exactly $m+n-1$ digits, since $\overbrace{111\ldots1}^{m \text{ ones}} \times \overbrace{111\ldots1}^{n \text{ ones}} < 2 \cdot 10^{m-1} \times 2 \cdot 10^{n-1} = 4 \cdot 10^{m+n-2}$ . If $m, n > 9$ , then considering the tenth leftmost digit, there would be a carry, thus the number consisting of the first nine digits of $N$ is bigger than the number formed by the last nine digits of $N$ , in reversed order. Then if $m$ and $n$ are bigger than $9$ the number $N$ is not a palindrome.

If one of the numbers $m$ , $n$ does not exceed $9$ then there won't be a carry and thus the number $N$ is palindrome.

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