Adv Alegbra HW Solutions 32

Adv Alegbra HW Solutions 32 - sum of its (decimal) digits...

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32 (iii) 243 x + 17 101 mod 725. Solution. The Euclidean algorithm gives 1 = 182 · 243 61 · 725. 243 x + 17 101 mod 725 gives 243 x 84 mod 725 . Hence x 182 · 84 = 15288 63 mod 725. (iv) 4 x + 3 4 mod 5. Solution. x 4 mod 5. (v) 6 x + 3 4 mod 10. Solution. 6 x + 3 4 mod 10 is the same problem as 6 x 1mod 10. There are no solutions. The candidates for x are all r with 0 r 9, and multiplying each of them by 6 never gives 1 mod 10. (Of course, ( 6 , 10 ) ±= 1.) (vi) 6 x + 3 1 mod 10. Solution. 6 x + 3 1 mod 10 is the same problem as 6 x 8mod 10. This congruence does have solutions. If 6 x 8 = 10 m , then 3 x 4 = 5 m , and so this gives a new congruence 3 x 4mod5 or x 8 3 mod 5. Thus, x = ... 2 , 3 , 8 , 13 ,... ; there are two possible solutions mod10, namely, x 3 mod 10 and x 8 mod 10. 1.79 Let m be a positive integer, and let m 0 be an integer obtained from m by rearranging its (decimal) digits (e.g., take m = 314159 and m 0 = 539114). Prove that m m 0 is a multiple of 9. Solution. By casting out 9s, a number is divisible by 9 if and only if the
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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 7-adic digits 202). Hence 10 100 3 100 3 4 = 81 4 mod 7 ....
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