assig6 - b we have a ³ ³ b if and only if a 2 ³ ³ b 2 ....

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MATH 135 Algebra, Assignment 6 Due: Wed Oct 28, 8:30 am 1: Find the prime factorization of each of the following integers. (a) 30! (b) ± 30 10 ² (c) 2 36 - 1 2: (a) Let a = 8400. Find the number of positive factors of a . (b) Find the number of positive integers whose prime factors are 2, 3 and 5 and which have exactly 100 positive divisors. (c) Let a = 6 Y k =1 k k . Find the number of factors (positive or negative) of a which are either perfect squares or perfect cubes (or both). 3: In parts (a) and (b), find the prime factorization of gcd( a, b ) and of lcm( a, b ). (a) a = 2 4 · 3 2 · 5 · 11 and b = 2 2 · 5 3 · 7 · 11 (b) a = 25! and b = (5500) 3 (1001) 2 . (c) Find the number of pairs of positive integers ( a, b ) with a b such that gcd( a, b ) = 60 and lcm( a, b ) = 4200. 4: (a) Show that for all positive integers a and
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Unformatted text preview: b we have a ³ ³ b if and only if a 2 ³ ³ b 2 . (b) Show that for all positive integers a , b and c , if c ³ ³ ab then c ³ ³ gcd( a, c ) gcd( b, c ). (c) Show that for all positive integers a and b we have gcd( a, b ) = gcd ( a + b, lcm( a, b ) ) . 5: A Hilbert number is a positive integer of the form n = 1 + 4 k for some integer k ≥ 0. A Hilbert prime is a Hilbert number n> 1 whose only Hilbert number factors are 1 and n . (a) List the first 20 Hilbert primes. (b) Show that every Hilbert number greater than 1 is either a Hilbert prime or a product of Hilbert primes. (c) Show that the factorization of a Hilbert number into Hilbert primes is not always unique....
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This note was uploaded on 05/04/2010 for the course MATH 135 taught by Professor Andrewchilds during the Fall '08 term at Waterloo.

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