Divisibility by 11: ProofDate: 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/ |
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