number theory
posted by Laura on .
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?

How do we know a number is divisible by 11?
e.g. is 164587467 divisible by 11?
Add up the oddnumber 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 evennumber positioned digits = 6+3+0+3 = 12
sum of oddnumber positioned digits = 1+4+7+0 = 12
now forming the palindrome would obviously make the original oddnumber positioned numbers into the evennumber 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