Sums Divisible by 11
Date: 10/10/2001 at 20:13:42 From: Chris Stava Subject: Why is the sum of an even-digit number and that same number in reverse always divisible by 11? Hi Dr. Math, You've been helpful before, so I'm calling on you again. A student's parent asked me why the sum of a number with an even number of digits (4, 6, 8...) and that same number written in reverse is always divisible by 11. (Example: 3456 + 6543) I know the tricks to test for divisiblity, but I don't know WHY. Can you help? I appreciate your time and effort. Thanks so much. Sincerely, Chris
Date: 10/10/2001 at 23:15:40 From: Doctor Peterson Subject: Re: Why is the sum of an even-digit number and that same number in reverse always divisible by 11? Hi, Chris. Suppose the number has digits "ABCD," so it and its reverse "DCBA" are 1000A + 100B + 10C + D and 1000D + 100C + 10B + A When we add them, we get 1001A + 110B + 110C + 1001D Since 110 = 11*10 and 1001 = 11 * 91, this is 11*91 A + 11*10 B + 10 C + 11*91 D This is clearly divisible by 11. How about larger numbers of even digits? Then we have [10^(2n-1) A + ... + 10^0 Z] + [ 10^(2n-1) Z + ... + 10^0 A] which becomes [10^(2n-1)+10^0]A + [10^(2n-2)+10^1]B + ... Each pair of opposite digits (A and Z, B and Y, and so on) has a multiplier of the form 10^(2n-1-k) + 10^k = 10^k (10^2(n-k)-1) + 1) Recall that x^(2n+1) + 1 (an odd power of x plus 1) can be factored as (x+1) times another polynomial with integer coefficients (since x = -1 is a zero); so 10+1 is a factor of (10^2(n-k)-1) + 1). This last part can be said more simply if you have to. Students may simply believe you when you say that the multipliers of pairs of digits will always be numbers like 11, 1001, 100001, and so on times a power of ten, and the former are all multiples of 11. See the Dr. Math FAQ and our archives on the reason for the divisibility rule for 11, which is closely related. Divisibility Rules - Dr. Math FAQ http://mathforum.org/k12/mathtips/division.tips.html Divisibility by 11 - Dr. Math archives http://mathforum.org/dr.math/problems/suria2.17.98.html - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/
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