C__work_280sol3 - Solutions to Homework 3 February 27, 2007...

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February 27, 2007 The following exercises are from Rosen. 2.4: 6 Suppose that a | c and b | d . Then there exists k , m such that c = ka , d = mb . That means cd = kmab , so ab | cd . 2.4: 12 (a) 39 = 3 · 13. (b) 81 = 3 4 . 2.4: 14 The number of zeros at the end of n is the largest k s.t. 10 k | n . Note that 10 k = 2 k 5 k . By the fundamental theorem of arithmetic, each i ∈ { 2 , 3 , . . . , 100 } can be uniquely represented as i = 2 a i 5 b i c i with a i , b i , c i N , and c i is a product of primes, none of which is 2 or 5. It follows that 100! = 2 100 i =2 a i × 5 100 i =2 b i × Π 100 i =2 c i Let k = min( i a i , i b i ). That is, k is the smaller of the sum of exponents of 2 and the sum of the exponents of 5. Then 100 = 10 k 2 ( 100 i =2 a i ) - k 5 ( 100 i =2 b i ) - k Π 100 i - 2 c i . Note that at least one of ( 100 i =2 a i ) - k and ( 100 i =2 b i ) - k is 0, so 2 ( 100 i =2 a i ) - k 5 ( 100 i =2 b i ) - k Π 100 i =2 c i is not a multiple of 10. (The exponent of at least on of 2 and 5 is 0.) Bottom line: there are k zeros at the end of 100!, where k is the smaller exponent of 2 and 5 when 100! is written as a product of primes. So now the problem is to compute k . There are 100 / 5 = 20 i s for which b i 1: the multiples of 5. Four of those twenty multiples are multiples of 25 and have b i = 2; the rest have b i = 1. Thus, 100 X i =2 b i = 1 · 16 + 2 · 4 = 24 . 1
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C__work_280sol3 - Solutions to Homework 3 February 27, 2007...

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