Divisibility by 9Date: 03/22/2002 at 08:18:34 From: Tonya Wagner Subject: Divisibility by 9 Explain why an integer that is rearranged and then subtracted from itself is always divisible by nine. I can show numerous examples and have looked through the math resources that I have, but I can not seem to come up with a good explanation. Date: 03/22/2002 at 08:41:16 From: Doctor Paul Subject: Re: Divisibility by 9 A number is divisible by nine if and only if the sum of the digits is divisible by nine. This follows from the fact that a number is divisible by three if and only if the sum of the digits is divisible by three and that a number is divisible by nine if and only if it is divisible by three twice. More about that here: Why do Divisibility Rules work? http://mathforum.org/k12/mathtips/division.tips.html Now, a permutation of the digits will not change the sum of those digits (since addition of real numbers is a commutative operation). Suppose that x = a - b, and that the number b is obtained by simply permuting the digits of a. Then a is either going to be: a multiple of nine (0 mod 9) 1 more than a multiple of nine (1 mod 9) 2 more than a multiple of nine (2 mod 9) 3 more than a multiple of nine (3 mod 9) 4 more than a multiple of nine (4 mod 9) 5 more than a multiple of nine (5 mod 9) 6 more than a multiple of nine (6 mod 9) 7 more than a multiple of nine (7 mod 9) 8 more than a multiple of nine (8 mod 9) Suppose that a is y more than a multiple of nine (a = y mod 9), where y = 0, 1, 2, ..., or 8. Then the sum of the digits of a will be y more than a multiple of nine. Since b is just a permutation of the digits of a, the sum of the digits of b will also be y more than a multiple of nine. Notice that in general, a - b will be divisible by nine if and only if they are congruent to the same number mod 9. I.e., if a is congruent to 7 mod 9, but b is congruent to 5 mod 9, then their difference will be congruent to (7 - 5) mod 9 = 2 mod 9 and hence will not be divisible by nine (a number that is two more than a multiple of nine is obviously not divisible by nine). Notice that what we have above is that regardless of the choice of a, that the sum of the digits of a will be the same as the sum of the digits of b and hence they will always be congruent to the same number mod 9. Thus their difference will be congruent to 0 mod 9 and hence will be divisible by nine. I hope this helps. Please write back if you'd like to talk about this some more. - Doctor Paul, The Math Forum http://mathforum.org/dr.math/ |
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