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Divisibility by 11: Proof


Date: 02/12/2002 at 13:29:48
From: Jenny
Subject: Proof with divisibility and alternating sums

Prove that a positive integer n is divisible by 11 if and only if the 
alternating sum of its digits is divisible by 11.

I'm not sure how to approach this proof. A hint I have is that 11 is 
congruent to -1 mod 10. Can you please help me?

Jenny


Date: 02/12/2002 at 13:42:51
From: Doctor Paul
Subject: Re: Proof with divisibility and alternating sums

   10^1 = -1 mod 11
   10^2 =  1 mod 11
   10^3 = -1 mod 11
   10^4 =  1 mod 11
   10^5 = -1 mod 11
   10^6 =  1 mod 11
   .
   .
   .

Now write out your number n digit by digit as follows:

   a_n ... a_3 a_2 a_1 a_0

So n = a_0 + a_1*10 + a_2*10^2 + a_3*10^3 + ...

Then take congruences mod 11

Therefore, if a_0 - a_1 + a_2 - a_3 + ... is divisible by 11, 
then so is n.

An example:

Take a number you want to check for divisibility by 11:

   1353 = 3 + 5*10 + 3*10^2 + 1*10^3

Now take congruences mod 11:

   1353 = 3 + 5*10 + 3*10^2 + 1*10^3

        = 3 + 5*(-1) + 3*1 + 1*(-1) mod 11

        = 3 - 5 + 3 - 1 = 0 

so 1353 is divisible by 11.

I hope this helps. Please write back if you'd like to talk about this 
more.

- Doctor Paul, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Number Theory

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