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Unformatted text preview: Homework 5 Solutions Kyle Chapman November 9, 2011 96.2 a Numbers n with exactly three positive divisors must be of the form p 2 for some prime number p . If at least two distinct prime factors p,q divided n , then 1 ,p,q,pq would be four distinct factors, so a number with three positive divisors must have only one prime factor, making n = p k for some k . If k < 2 then there are fewer than three factors, and if k > 2 then there are more than three factors. b Numbers n with exactly four prime factors must either be p 3 for some prime p or pq for distinct primes p,q . If there were three distinct prime factors p,q,r then 1 ,p,q,r,pq would all be prime factors so any number must have no more than four prime factors. If it has two prime factors but the power on one of them is greater than 1, then without loss of generality it can be written as p k q j where k > 1 and we get that 1 ,p,q,pq,p 2 are all factors, so if n has two prime factors, their exponents must both be one. The last case is if n = p k for some k , and we can see that only k = 3 works. c First, we note that any number which is at least 2 can be written as 2 s + 3 r where both s,r are non negative. The first few cases give 2 = 2(1) + 3(0) , 3 = 2(0) + 3(1). After this point, if n = 2( r ) + 3( s ) with r,s nonnegative, then either n + 1 = 2( r 1) + 3( s + 1) or r was 0, which means s ≥ 1 since n ≥ 3 so n + 1 = 2( r + 2) + 3( s 1). Thus, in either case, we get the inductive step and so the result holds. Now we know that our number n has a prime factorization n = Π n i =1 p q i i . We can assume each q i > 0 and so each p i  n so...
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 Fall '08
 Bigelow
 Factors, Prime number, Divisor, prime factors, Kyle Chapman

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