Integer DivisorsDate: 04/11/97 at 21:34:00 From: Mafaza Mahroof Subject: Integer Divisors The positive integer N has exactly six distinct integer divisors including 1 and N. The product of five of these is 648. Which one of the following integers must be another divisor of N? A. 4 B. 9 C. 12 D. 16 E. 24 I don't get the question and can you tell me step by step you got the answer? Thanks very much for the help, Mafaza Date: 04/12/97 at 13:32:30 From: Doctor Anthony Subject: Re: Integer Divisors A number like 24 has prime factors 2^3 x 3 To find the number of distinct divisors, including 1 and 24, we can take the factor 2 0, 1, 2, 3 times = 4 choices, factor 3 0, 1 times = 2 choices. So altogether there will be 4 x 2 = 8 distinct factors. To check if this is correct: 1, 2, 3, 4, 6, 8, 12, 24 = 8 factors, so it is correct. We use these ideas to answer the given problem. 648 is divisible by both 2 and 3 so we know that both will be prime factors of the number N. So N = 2^2 x 3 or N = 2 x 3^2. Both of these would give 6 distinct factors including 1 and N. If N = 2^2 x 3 then the factors are 1, 2, 3, 4, 6, 12. No product of 5 of these gives 648. So try N = 2 x 3^2 with factors 1, 2, 3, 6, 9, 18. Now 1 x 2 x 3 x 6 x 18 = 648, and thus five of the factors give 648 as required. We can conclude that N = 18. Since we know N = 18, the only number that must be a factor from those listed (4, 9, 12, 16, 24) is 9. -Doctor Anthony, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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