Show every dec. palind. int. that has an even # of digits is div. by 11.
18833881 is divisible by 11; why?
Every 2 digit palindrome is obviously divisible by 11.
Suppose every 2n digit palindrome is divisible by 11.
Consider any 2(n+1) digit palindrome P. The central 2n digits define a 2n
digit palindrome PP divisible by 11, so write:
P=PP*10 + (10^(2n+1)+1)Q
where Q is the most (and least) significant digit of P. Then P is divisible by
11 because PP is and so is 10^(2n+1)+1.
Then the proof follows by mathematical induction.
(You may feel the need to prove that 10^(2n+1)+1 is divisible by 11.
This would proceed along the lines of observing that 10=-1 mod 11
so any odd power of 10 is also congruent to -1 modulo 11, which is
equivalent to any odd power of 10 plus 1 is divisible by 11)
The simplest way to see this is by using the rule that to determine if a number is divisible by 11 if and only if the "alternating sum" is divisible by 11.
Thus, take for example 12344321
The alternating sum is,
And 0 is divisible by 11.
In fact, any even palindrom is divisible by 11 because the alternating sum (from right to left) is going to be zero.
Thus, if a number is an ODD palindrome with zero as its middle then it too will be divisible by 11. For example, 12021 then alternating sum is,
Furthermore, the middle digit in an odd palindrome determines the the remainder upon division (excerise).