Manhattan Question

This topic has expert replies
Master | Next Rank: 500 Posts
Posts: 120
Joined: Thu May 15, 2008 1:07 pm
Location: Boston .US
Thanked: 1 times

Manhattan Question

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.
Source: — Problem Solving |

Senior | Next Rank: 100 Posts
Posts: 87
Joined: Tue Jul 15, 2008 10:23 am
Location: Lima
Thanked: 4 times
Followed by:1 members

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

x+3y<1000

2x2x2x2x2x2x2x2x2 + 3x2x2x2x2x2x2x2 = 896

What's the OA?

Master | Next Rank: 500 Posts
Posts: 377
Joined: Wed Sep 03, 2008 9:30 am
Thanked: 15 times
Followed by:2 members

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.

Senior | Next Rank: 100 Posts
Posts: 35
Joined: Mon Jun 23, 2008 1:41 pm
Thanked: 2 times

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?

Master | Next Rank: 500 Posts
Posts: 120
Joined: Thu May 15, 2008 1:07 pm
Location: Boston .US
Thanked: 1 times

by priyankamishra11 » Sat Sep 06, 2008 6:32 pm
OA is 16.
Regards,
Priyanka