# Thread: Decimal Palindromic Integer

1. ## Decimal Palindromic Integer

Show every dec. palind. int. that has an even # of digits is div. by 11.

For example,

18833881 is divisible by 11; why?

2. Originally Posted by Ideasman
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)

3. Originally Posted by Ideasman
Show every dec. palind. int. that has an even # of digits is div. by 11.

For example,

18833881 is divisible by 11; why?
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).

4. Originally Posted by ThePerfectHacker
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.
Uh, how? 12021 is not divisible by 11.

It should be:

1 - 2 + 0 - 2 + 1 = -2, which does not divide 11.

5. Originally Posted by Ideasman
Uh, how? 12021 is not divisible by 11.

It should be:

1 - 2 + 0 - 2 + 1 = -2, which does not divide 11.
Okay, relax, thus I made a mistake.