Math245Sols3Sum00

# Math245Sols3Sum00 - MATH 245: QUIZ 3 SOLUTIONS a) b) 980 =...

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MATH 245: QUIZ 3 SOLUTIONS 1) a) 980 = (2)(2)(5)(7)(7) = 2 2 5 7 2 b) 616 = (2)(2)(2)(7)(11) = 2 3 7 11 c) The prime factors of interest are 2, 5, 7, and 11. For the lcm, we take the larger exponent on each prime factor. 980 = 2 2 5 1 7 2 11 0 616 = 2 3 5 0 7 1 11 1 lcm = 2 3 5 1 7 2 11 1 = 21,560 d) For the gcd, we take the smaller exponent on each prime factor. 980 = 2 2 5 1 7 2 11 0 616 = 2 3 5 0 7 1 11 1 gcd = 2 2 5 0 7 1 11 0 = 2 2 7 = 28 2) Let a = 4743 and b = 867. ab = gcd a , b ( ) lcm a , b ( ) 4743 ( ) 867 ( ) = 51 ( ) a , b ( ) a , b ( ) = 4743 ( ) 867 ( ) 51 = 80,631 3) Let n be a composite integer. Then, n = rs for some integers r and s strictly between 1 and n . If both r and s are greater than , then rs > n n = n , which contradicts " n = rs ." So, one or both of r or s must be less than or equal to . 4) Think of 100! as (1)(2)(3) (100). We want to count the number of "3-factors" in the integers from 1 to 100. There are 100 3 = 33 multiples of 3 between 1 and 100. (3, 6, 9, …, 99) We can pick up one 3-factor from each of these 33 multiples.

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## This note was uploaded on 09/08/2011 for the course MATH 245 taught by Professor Staff during the Spring '11 term at Mesa CC.

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Math245Sols3Sum00 - MATH 245: QUIZ 3 SOLUTIONS a) b) 980 =...

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