How do I judge if a positive integer is divisible by 11?

Question

If possible, give me a math statement and use binomial theorem to prove it. Or just give me a link to a site

Answer 1

I can't find anything using binomial theorem to prove it online, but Wikipedia has a couple methods: http://en.wikipedia.org/wiki/11_(number)...

I know it's there, I saw it a long time ago, but I haven't seen a proof, and my mathematical proofs are way too rusty. Here's what I see, at least for 3-digit numbers, maybe you can take it from here.

Convert the number to a polynomial, where x = 10.

121 = 100 + 20 + 1 = x^2 + 2x + 1 = (x+1)(x+1); Since x+1 = 11, 121 is divisible by 11

132 = x^2 + 3x + 1 = (x+2)(x+1); divisible by 11

144 = x^2 + 4x + 4 = (x+2)(x+2); not divisible by 11

100 = x^2 + 0x + 0 = (x+0)(x+0); not divisible by 11

88 = 0x^2 + 8x + 8 = (0x+8)(x+1); divisible by 11

Edit: Actually, that would work for all numbers by converting it to a polynomial where x = 10. If (x+1) is a factor, it's divisible by 11, I just can't remember how to prove it.

Edit: the answer below mine works for all numbers from 1-999. For all two digit numbers divisible by 11 below 100  a = 0 and b = c, so a + c =b. For 10, 0 + 0 <> 1. For 1-9, 0 + c <> 0.

Edit: Re: the answer below mine again. It's just a simplified variation of the first method on the Wikipedia link.

Answer 2

Here's a really quick way for 3 digit numbers. (I think this works for all cases....)

If the number is, say;

'abc', where each letter represents an integer (the numbers are not multiplied together)

'abc' is divisible by 11 if a+c=b.