Integer Divisors

Date: 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/