Which of the following is an integer? I. 12!/6! II. 12!/8! : Quant Question Archive [LOCKED]
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# Which of the following is an integer? I. 12!/6! II. 12!/8!

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Which of the following is an integer? I. 12!/6! II. 12!/8! [#permalink]

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27 Nov 2006, 16:47
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Which of the following is an integer?

I. 12!/6!

II. 12!/8!

III. 12!/7!5!

A) I ONLY
B) II ONLY
C) III ONLY
D) I AND II
E) I, II & III

What is a quick way to solve these typese of questions. Thanks
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27 Nov 2006, 16:57
Quote:
What is a quick way to solve these typese of questions. Thanks
Good question, does anyone know?
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27 Nov 2006, 17:01
You can see that I&II are integers because the denominator factorials just cancel out part of the numerator factorials.

In III if you cancel out 7! leaving 12*9*8/5! then 5 and 3 are no longer factors of the numerator and so it wont yield an integer.
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27 Nov 2006, 17:30
It should be E
12! / 7! * 5! = (12 * 11 * 10 * 9 * 8) / (5 * 4 * 3 * 2) = 11 * 9 * 8
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27 Nov 2006, 17:33
Think I may have had a brain fart on that one:

III gives: 12*11*10*9*8/5!

10 = 2*5 and 12 = 3*4 thus containing the prime factorisation of 5!

E it is
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27 Nov 2006, 18:00
Thanks guys - E it is. Am I correct to assume that since all choices have the prime factorization of five in numerator and denominator that these will yield an integer?
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27 Nov 2006, 18:06
It's because all the numerators contain ALL the factors of the denominators.

In the last example:

12! / 7! * 5! = (12 * 11 * 10 * 9 * 8) / (5 * 4 * 3 * 2) = 11 * 9 * 8

The "12" is cancelled out by the (4*3) and the "10* is cancelled out by the (5*2) leaving 11*9*8
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Re: GMAT Prep - ! [#permalink]

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28 Nov 2006, 08:11
[quote="Matrix02"]Which of the following is an integer?

I. 12!/6!

II. 12!/8!

III. 12!/7!5!

A) I ONLY
B) II ONLY
C) III ONLY
D) I AND II
E) I, II & III

12!/6! will be intger
12!/8! will be integer
12*11*10*9*8*6/5*4*3*2= integer
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28 Nov 2006, 09:02
It's E,the first two options can be stratifyed straight forward but the third can also be satisfyed as 12!/7!5!=12*11*10*9*8/5*4*3*2*1
here 5!=120=10*12
hence,it can be written as 12*11*10*9*8/10*12 and therefore it must be a integer..
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28 Nov 2006, 12:05
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28 Nov 2006, 12:05
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