Three digit palindromes

Three digit palindromes

What is the smallest three digit palindrome divisible by 18?  Can you solve this without a brute force examination of possibilities?

Communicated by Tom Chiari.

 

FIRST THOUGHTS

We approach this problem by first thinking about what it means for a number to be divisible by 18. For a number to be divisible by 18, it is both necessary and sufficient that it be divisible by both 9 and 2. What will that imply?

 

SECOND THOUGHTS

Since the palindrome we are seeking is even, it must end with (and thus start with) one of the digits 0, 2, 4, 6, 8. Under most definitions of three digit numbers, 0 cannot be the leading digit, so our next smallest candidate is 2. This is the smallest possible leading digit for our palindrome.

 

THIRD THOUGHTS

Since our palindrome is also divisible by 9, we know by the divisibility rule for multiples of 9 that its digits must add up to a multiple of 9. A reader who is unfamiliar with this rule can see evidence for it (though not a proof!) when considering 9 * 3 = 27, where 2 + 7 = 9, and 9 * 4 = 36, where 3 + 6 = 9.

We know that our palindrome begins and ends with 2, so we see that 2 + middle digit + 2 = 9. It follows that middle digit = 5. Our palindrome is thus 252.

 

NEXT STEPS

Try and prove the divisibility rule for 3; see https://artofproblemsolving.com/wiki/index.php?title=Divisibility_rules/Rule_for_3_and_9_proof

For the brave: extend and find rules for 11 and some other numbers (some like 2 and 5 are easy, 4 and 8 a little more work, …).