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by yellowho » Fri Jan 28, 2011 9:27 pm
If p is a positive integer and 10p/96
is an integer, then the minimum
number of prime factors p could have is

(A) One
(B) Two
(C) Three
(D) Four
(E) Five

I had 5. OA is 2. It didn't say different. How is everyone else interpreting this question.
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by schumi_gmat » Fri Jan 28, 2011 11:49 pm
96 = 2^5 * 3

10p/96 = 2*5*p/96

for above equation to be an integer p = 2^4 *3

There are 2 prime factors - 2 and 3
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by Anurag@Gurome » Sat Jan 29, 2011 12:41 am
yellowho wrote:If p is a positive integer and 10p/96 is an integer, then the minimum number of prime factors p could have is

(A) One
(B) Two
(C) Three
(D) Four
(E) Five
10p/96 = 5p/48 ---> An integer

For 5p/48 to be integer, p has to be a multiple of 48 = (2^4)*(3)

Hence, p must have at least 2 prime factors, i.e. 2 and 3.

The correct answer is B.
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by prachich1987 » Thu Feb 03, 2011 3:09 am
Anurag@Gurome wrote: 10p/96 = 5p/48 ---> An integer

For 5p/48 to be integer, p has to be a multiple of 48 = (2^4)*(3)

Hence, p must have at least 2 prime factors, i.e. 2 and 3.

The correct answer is B.
The question is not asking us to find DISTINCT prime factors.
It is just asking us to find prime factors.
The minimum no. of prime factors should be 5.

Can you please explain.
Thanks!
Prachi
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