How many 3 digit palindromes are divisible by 9?

Interesting problem.

A palindrome is a number which is identical when read/interpreted left-to-right or right-to-left.   For example, 14241 is a palindrome, so is 444444.

A three digit palindrome must be made up of two distinct digits only, one for the first and last digits, and one for the middle, for example, 424, or 515, etc.

For a number to be divisible by 9, the sum of the digits must add up to a multiple of 9, such as 0, 9, 18, 27, etc.

We can now form a table of the possible 3-digit palindromes divisible by 9, starting with the first/last digits (=A), and finding the middle digit (=B).  The number is therefore represented by the pattern ABA.

For example, when A=1, we have the number 1B1.  For the number to be divisible by 9, B=7 so that sum=1+7+1=9, which is divisible by 9.

For each value of A (from 1 to 9) we can only find one value of B that makes the number divisible by 9.   Thus we can only find nine such numbers, with A equal to 1 to 9.

Here's a list of the numbers

A   B   number

1    7    171

2   5    252

3   3    333

4   1     414

5   8    585

6   6    666

7   4    747

8   2   828

9   0   909

kaahe btau