Adv Alegbra HW Solutions 31

Adv Alegbra HW Solutions 31 - n with n 1 mod 100 and n 4...

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31 c = b ` = db 0 ` , so that a 0 m = b 0 ` . Thus, a 0 divides b 0 ` ;a s ( a 0 , b 0 ) = 1, we have a 0 divides ` , by Corollary 1.40. Write ` = a 0 k , and observe that c = db 0 ` = db 0 a 0 k = ( db 0 )( da 0 ) k / d =[ ab / d ] k . Therefore, ab / d =[ a , b ] , and so [ a , b ] ( a , b ) = ab . (ii) Find [1371, 123]. Solution. [ 1371 , 123 ]= 1371 · 123 /( 1371 , 123 ) . By the Eu- clidean algorithm, ( 1371 , 123 ) = 3, and so [ 1371 , 123 ]= 56 , 211. 1.77 True or false with reasons. (i) If a and m are integers with m > 0, then a i mod m for some integer i with 0 i m 1. Solution. True. (ii) If a , b and m are integers with m > 0, then a b mod m implies ( a + b ) m a m + b m mod m . Solution. False. (iii) If a is an integer, then a 6 a mod 6. Solution. False. (iv) If a is an integer, then a 4 a mod 4. Solution. False. (v) 5263980007 is a perfect square. Solution. False. (vi)
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Unformatted text preview: n with n 1 mod 100 and n 4 mod 1000. Solution. False. (vii) There is an integer n with n 1 mod 100 and n 4 mod 1001. Solution. True. (viii) If p is a prime and m n mod p , then a m a n mod p for every natural number a . Solution. False. 1.78 Find all the integers x which are solutions to each of the following congru-ences: (i) 3 x 2 mod 5. Solution. x 4 mod 5. (ii) 7 x 4 mod 10. Solution. x 12 mod 10....
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