2017/18 AHSMC Part 1 Problem 15

A five digit palindrome has the form $abcba = 10001a + 1010b + 100c$ with $a \ne 0$. In order for the number to be divisible by $5$, we must have $a = 5$. Therefore, the number is $50005 + 1010b + 100c$.

Since $50005 + 1010b + 100c = 11(4546 + 92b + 9c) - 1 - 2b + c$, we then need to have $c - 2b - 1$ divisible by $11$. Since $-19 \le c - 2b - 1 \le 8$ we must have $c - 2b - 1 = 0$ or $c - 2b - 1 = -11$.

If $c - 2b - 1 = 0$ then $c = 2b + 1$ which leads to $5$ pairs $(b,c) \in \{(0,1),(1,3),(2,5),(3,7),(4,9)\}$.

If $c - 2b - 1 = -11$ then $2b = c + 10$ and hence $b \ge 5$. Then we get $(b,c) \in \{(5,0),(6,2),(7,4),(8,6),(9,8)\}$.

Therefore we have $10$ possibilities. The answer is (C).