1997MDate: 08/12/2001 at 06:59:55 From: Brian Hayne Subject: Integers and divisors Hey Dr Math! I need some help working out the answer to this question: Find all composite positive integers M such that the product of 1997 and M has exactly four divisors. Date: 08/13/2001 at 19:19:17 From: Doctor Greenie Subject: Re: Integers and divisors Hi, Brian - There is only one. The number of divisors of a number is directly related to the prime factorization of the number. If you are not familiar with this fact, try searching the Dr. Math archives from http://mathforum.org/mathgrepform.html using the exact phrase number of divisors . Follow one or more of the links that search provides you. If a number has exactly four divisors, then there are only two possibilities for the prime factorization: (1) The number is the product of two different prime numbers p and q; the divisors are {1, p, q, pq}. (2) The number is the cube of a prime number p; the divisors are {1, p, p^2, p^3}. The number 1997 is a prime number. If 1997M has exactly four divisors and case (1) above holds, then M must be a prime number other than 1997. But the problem specifies that M be a composite number. So there are no solutions of this type. If 1997M has exactly four divisors and case (2) above holds, then 1997M must be the cube of a prime number. Since 1997 is a prime number, the only way for 1997M to be a cube is for M to be equal to 1997^2. - Doctor Greenie, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2015 The Math Forum
http://mathforum.org/dr.math/