Posted by **Laura** on Sunday, April 19, 2009 at 2:53pm.

How do you prove that any palindrome, a number that reads the same backwards and forwards, with an even number of digits is divisible by 11?

I know how to show a number is divisible by 11, but how do you prove the above case?

- number theory -
**Reiny**, Sunday, April 19, 2009 at 3:19pm
How do we know a number is divisible by 11?

e.g. is 164587467 divisible by 11?

Add up the odd-number positioned digits

1+4+8+4+7 = 24

6+5+7+6 = 24

if their sum has a difference of zero or a multiple of 11, then the original number divides by 11

e.g. 16437003

sum of even-number positioned digits = 6+3+0+3 = 12

sum of odd-number positioned digits = 1+4+7+0 = 12

now forming the palindrome would obviously make the original odd-number positioned numbers into the even-number positioned digits and vice versa.

so their sum would still have either a difference of zero or a multiple of 11.

BTW, the same property would result if you formed the palindrome of a number with an even number of digits

e.g. 1737032 and 2307371 are palidromes, and both are divisible by 11

the sum of the digits of their respective groupings are 6 and 17

and the difference between these two is 11

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