Tough one- difficult to solve in 2 min

[ankur.agrawal]

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

[HSPA]

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

[force5]

Hi IMO- D (16)

2^9 = 512+ 3*2^7 <1000
hence 9+7=16

[HSPA]

hello force5,

How did you get 16?

[force5]

one factor is already given as 3 after counting 2^9

[HSPA]

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

[force5]

thanks HSPA you just missed on one factor.

hope it helped