Palindromic Numbers

We know that any four digit palindromic number is divisible by 11. To see that take the following:

$abba = a\times 1000+ b\times 100+b\times 10 +a = 1001a+ 110b= 11(91a+10b)$ , which indicates that $abba$ is divisible by $11$ .

In general, any palindromic number with even number of digits is divisible by 11.

Now, we need to take out of those the ones that are divisible by 9 too.

$9| abba$ $\implies 9|2(a+b)$ $\implies 9|(a+b)$

Since $a \neq 0$ $\implies a+b=9 \ or\ a+b=18$ .

For $a+b=9$ , we have 9 solutions $(1,8),(2,7),(3,6),...,(9,0)$ and for $a+b=18$ we have 1 solution $(9,9)$ .

Thus, the number of 4-digit palindromic numbers that are divisible by (11 and 9) is 10.