— Burt Kaliski Jr. (@modulomathy) June 20, 2012
The sum of the digits of a number divisible by 9 is itself divisible by 9; this follows from the fact that powers of 10 are always one more than a multiple of 9.
For the solution, let’s look at one-digit, two-digit, and three-digit numbers separately.
- One digit: All one-digit numbers are palindromes; the only one divisible by 9 is 9.
- Two digits: The only two-digit numbers that are palindromes are the ones divisible by 11. The only one also divisible by 9 is 99.
- Three digits: Any three-digit number of the form aba is a palindrome. The only ones divisible by 9 are those for which a+b+a = 2a+b is divisible by 9. With the limitation to day-numbers, we know that a must be 1, 2 or 3. This limits the answers to 171, 252, and 333.
Answer: 9, 99, 171, 252, 333.