Numeric palindromes

[sinsi]

Something on the back page of our Sunday paper.

All palindromes with an even number of digits are divisible by 11

I assume he means "evenly divisible".

Just a bit of trivia that I found interesting...now to write the program to test it.

[Mark Jones]

Here is an easy way to test if a number (three digits or larger) is divisible by 11:

176 - this number is... 1+6 = 7 and 7 = 7
3542 - this number is... 3+4 = 7 and 5+2 = 7
92081 - this number is... 9+0+1 = 10 and 2+8 = 10
123456 - this number is not... 1+3+5 = 9   !==  2+4+6 = 12, however
123453 - this number is... 1+3+5 =9, and 2+4+3 = 9

[raymond]

Stated differently, it means that all palindromes with an even number of digits are multiples of 11. And you don't have to write a program to test that. As Mark Jones indicated, if the sum of every second digit is the same as the sum of the remaining digits (or if their difference is a multiple of 11), the number is a multiple of 11.

Consider the following palindrome where you can replace any of the letters by whatever digit you want:

ABCDEFFEDCBA

the sum of every 2nd digit is: A+C+E+F+D+B = A+B+C+D+E+F
and the sum of the remaining digits is: B+D+F+E+C+A = A+B+C+D+E+F

Therefore .......