Unformatted text preview: sum of its (decimal) digits is divisible by 9. But m and m have the same set of digits, for one is just a permutation of the other, and so the sum of their digits is the same. Hence, one is divisible by 9 if and only if the other one is. 1.80 Prove that a positive integer n is divisible by 11 if and only if the alternating sum of its digits is divisible by 11. Solution. Since 10 ≡ − 1 mod 11, a = d k 10 k + ··· + d 1 10 + d ≡ d k ( − 1 ) k + ··· − d 1 + d . 1.81 What is the remainder after dividing 10 100 by 7? Solution. Use Corollary 1.67 after noting that 100 = 2 · 7 2 + 2 (of course, this says that 100 has 7adic digits 202). Hence 10 100 ≡ 3 100 ≡ 3 4 = 81 ≡ 4 mod 7 ....
View
Full
Document
This note was uploaded on 12/21/2011 for the course MAS 4301 taught by Professor Keithcornell during the Fall '11 term at UNF.
 Fall '11
 KeithCornell

Click to edit the document details