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Tough one- difficult to solve in 2 min

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by ankur.agrawal » Tue Mar 22, 2011 8:34 am
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?

5

6

15

16

18.

OA after some discussion
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by HSPA » Tue Mar 22, 2011 8:47 am
here we have a clue... not necessarly distinct... so maximum count can come if most factor are 2

Let x = 2^8 = y then x+3y = 4(2^8) ~=1000

So 8+8 = 16 ... we need a little less

IMO 15 is the answer

This is what all I got in <2min
Last edited by HSPA on Tue Mar 22, 2011 8:48 am, edited 1 time in total.
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by force5 » Tue Mar 22, 2011 8:48 am
Hi IMO- D (16)

2^9 = 512+ 3*2^7 <1000
hence 9+7=16
Last edited by force5 on Tue Mar 22, 2011 8:52 am, edited 2 times in total.
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by HSPA » Tue Mar 22, 2011 8:50 am
hello force5,

How did you get 16?
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by force5 » Tue Mar 22, 2011 8:53 am
one factor is already given as 3 after counting 2^9
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by HSPA » Tue Mar 22, 2011 8:57 am
bravo force5...

options are a little tough for me .. it should be at least 10,16,25....
Looks more of an IIM questiion than a GMAT
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by force5 » Tue Mar 22, 2011 8:58 am
thanks HSPA you just missed on one factor.

hope it helped
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