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Manhattan Question

Expert replies
by priyankamishra11 » Fri Sep 05, 2008 8:39 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

Is there any short way to solve this question, or i have to just put all values and check it? .. Its very time consuming.
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Source: — Problem Solving |

by Fab » Fri Sep 05, 2008 11:58 am
I would go with 16:

x+3y<1000

2x2x2x2x2x2x2x2x2 + 3x2x2x2x2x2x2x2 = 896

What's the OA?
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by schumi_gmat » Fri Sep 05, 2008 12:04 pm
My Answer is 15

What is OA?

max value is y = 332 for x+3y<1000 to be true.

The length can be increased if we have lowest divisor. The lowest divisor is 2.

hence we can have 2^8 = 256

and x = 2^7 = 128


hence length is 15.
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by mayur00 » Fri Sep 05, 2008 1:27 pm
I used an approached similar to Fab's i.e. maximize 2's. However I think there is a slight typo in his post
2x2x2x2x2x2x2x2x2 + 3x2x2x2x2x2x2x2 = 896

The extra 3 should go and the answer should be 16. What is the OA?
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by priyankamishra11 » Sat Sep 06, 2008 6:32 pm
OA is 16.
Regards,
Priyanka
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