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Complex PS problem

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by Auzbee » Mon Dec 31, 2007 6:59 pm
For any integer k > 1, the term “length of an integer” refers to the number of positive prime factors, not necessarily distinct, whose product is equal to k. For example, if k = 24, the length of k is equal to 4, since 24 = 2 × 2 × 2 × 3. If x and y are positive integers such that x > 1, y > 1, and x + 3y < 1000, what is the maximum possible sum of the length of x and the length of y?

a. 5
b. 6
c. 15
d. 16
e. 18

OA is D. I am looking for a quick explanation to crack this. Not through the long method of solving it.
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by gmatguy16 » Sun Jan 06, 2008 1:52 pm
i am not sure whats the quickest method but its obvious that we need minimum prime factors to get maximum value..lets start with 2 ..
2^10 = 1024 >1000 ,so lets use 2 ^ 9 = 512..we still have 1000-512=488...
488/3 = 162..... so y can be 2 ^7 = 128..hence answer is 9+7 = 16
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by StarDust845 » Tue Jan 08, 2008 8:15 am
gmatguy16 wrote:i am not sure whats the quickest method but its obvious that we need minimum prime factors to get maximum value..lets start with 2 ..
2^10 = 1024 >1000 ,so lets use 2 ^ 9 = 512..we still have 1000-512=488...
488/3 = 162..... so y can be 2 ^7 = 128..hence answer is 9+7 = 16
I think it's a great way to solve this problem.

Calista.
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