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500 PS # 15

by dunkin77 » Sun Apr 01, 2007 6:07 am
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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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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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! :D
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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! :D
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!
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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? :)
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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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by Jeff@TargetTestPrep » Sun Jun 24, 2018 5:01 pm


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
Since 77 = 7 x 11, we see that k can be 2 x 3 x 5 x 13, or any factor of 2 x 3 x 5 x 13. Since 26 = 2 x 13, B is the correct answer.

Answer: B

Jeffrey Miller
Head of GMAT Instruction
[email protected]

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