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dunkin77: Master | Next Rank: 500 Posts; Posts: 269; Joined: Sun Apr 01, 2007 5:41 am

500 PS # 15

by dunkin77 » Sun Apr 01, 2007 6:07 am

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Hi,

The answer is B) but I would like to get explanation - can anyone help?

n= 2*3*5*7*11*13 / 77K

If n is an integer and then which of the following could be the value of k?
(A) 22
(B) 26
(C) 35
(D) 54
(E) 60

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Source: Beat The GMAT — Problem Solving | Sun Apr 01, 2007 6:07 am

ajith: Legendary Member; Posts: 1275; Joined: Thu Sep 21, 2006 11:13 pm; Location: Arabian Sea; Thanked: 125 times; Followed by:2 members

Re: 500 PS # 15

by ajith » Sun Apr 01, 2007 6:58 am

dunkin77 wrote:Hi,

The answer is B) but I would like to get explanation - can anyone help?

n= 2*3*5*7*11*13 / 77K

If n is an integer and then which of the following could be the value of k?
(A) 22
(B) 26
(C) 35
(D) 54
(E) 60

n= 2*3*5*7*11*13 / 77K
= 2*3*5*7*11*13/ (11*7)K
= 2*3*5*13/k

For this to be an integer K should be a multiple of 2,3,5,13 or a combination of these

22 = 11*2 -Violates the condition
26 = 2*13 (perfect!)
35 = 7 * 5- Violates the condition
54 = 3*3*3*2 -- Violates the condition
60 = 3*2*2*5 - Violates the condition

Always borrow money from a pessimist, he doesn't expect to be paid back.

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dunkin77: Master | Next Rank: 500 Posts; Posts: 269; Joined: Sun Apr 01, 2007 5:41 am

by dunkin77 » Sun Apr 01, 2007 9:02 am

Thanks,

Yes, I figured since, 11*7 is already included in 77K, K must be the combination of 2,3,5,13. So, the combnation dose not necesarily include all of them but just need to include part of them without including other numbers, corrrect?

Thanks again!

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ajith: Legendary Member; Posts: 1275; Joined: Thu Sep 21, 2006 11:13 pm; Location: Arabian Sea; Thanked: 125 times; Followed by:2 members

by ajith » Sun Apr 01, 2007 9:18 am

dunkin77 wrote:Thanks,

Yes, I figured since, 11*7 is already included in 77K, K must be the combination of 2,3,5,13. So, the combnation dose not necesarily include all of them but just need to include part of them without including other numbers, corrrect?

Thanks again!

You are right

K must be the combination of 2,3,5,13 and when you factorize K it should contain these numbers (2,3,5,13) only once.
It can be part or full , no problem!

Always borrow money from a pessimist, he doesn't expect to be paid back.

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aim-wsc: Legendary Member; Posts: 2469; Joined: Thu Apr 20, 2006 12:09 pm; Location: BtG Underground; Thanked: 85 times; Followed by:14 members

by aim-wsc » Sun Apr 01, 2007 9:55 am

wow! Alith is back 8)
go to intro section and tell us how have you been?

Getting started @BTG?
Beginner's Guide to GMAT | Beating GMAT & beyond
Please do not PM me, (not active anymore) contact Eric.

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Cybermusings: Legendary Member; Posts: 559; Joined: Tue Mar 27, 2007 1:29 am; Thanked: 5 times; Followed by:2 members

by Cybermusings » Mon Apr 02, 2007 5:14 am

77 is a product of 7 and 11 so now you have the equation 2*3*11*13 / k left for you

For 2*3*11*13 / k to be an integer it is important that the product of two or more of the set (2,3,11, 13) be completely divisible by k. Now look at Choice B i.e. 26 ;
26 = 13*2 ; Hence it's possible for k = 26

This is not possible for any other number

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