Show every dec. palind. int. that has an even # of digits is div. by 11.
For example,
18833881 is divisible by 11; why?
Proof is by induction.
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.
RonL
(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)
This problem appeared as an excerise in my number theory book about 3 years ago, since it was such an interesting fact, I memorized it ever since.
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,
1-2+3-4+4-3+2-1=0
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,
1-2+0+2-1=0.
Furthermore, the middle digit in an odd palindrome determines the the remainder upon division (excerise).