Consider three-digit numbers of the form 'aba' (first and last digits equal), such as 121, 343, 595.
How many such three-digit numbers are divisible by 5?

For 'aba' to be divisible by 5, the last digit a must be 0 or 5. But a is also the first digit, which cannot be 0 in a three-digit number, so a = 5. Then the middle digit b can be any of 0 to 9, giving 505, 515, ..., 595 — that is 10 numbers.