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Re: Which of the following is an integer? [#permalink]
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NickTW wrote:
Did a search, could not find anything on this one (the question stem is pretty vague). Would appreciate some enlightenment on the shortcut I'm missing!

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 only
E) I, II, and III

Source: GMAT Prep Exam. Solution : E



To me the question looks straight forward.
we know 12! is multiple of all numbers from 1 to 12, so dividing it by 6! or 8! is pretty much sure an integer.

now coming to the 3rd options 12! by 5! leaves 12*11*10*9*8*7*6/1*2*3*4*5*6*7 which results in integer, so the answe E is right
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Re: Which of the following is an integer? [#permalink]
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There is no shortcut persay...

Clear E.



12!/6! = 12*11*10*9*8*7*6!/6! = Integer
12!/8! = 12*11*10*9*8!/8! = Integer
12!/7!*5! = 12*11*10*9*8/5*4*3*2*1 = integer

So all 3 are correct.
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Re: Which of the following is an integer? [#permalink]
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Every factorial is divisible by every smaller factorial, so 12!/6! and 12!/8! are integers. 12!/(7!*5!) is the number of ways to choose 7 things from a group of 12 if order is irrelevant, so it must also be an integer.
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Re: Which of the following is an integer? [#permalink]
case 1 and case 2 are simple but case 3 is harder. for case 3, we have to write down all the factors.
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Re: Which of the following is an integer? [#permalink]
NickTW wrote:
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 only
E) I, II, and III


For III, I went through and crossed out everything that's reduced to visualize...
12*11*102*93*8*7*6*5*4*3*2*1 / 7*6*5*4*3*2*1 * 5*4*3*2*1

So you're left with 12*11*2*3 (from taking 5 and 3 from 10 and 9) = 792. Makes sense.

ScottTargetTestPrep
As a rule, you said m = n + p, then m!/(n!p!). When does the factorial in the denominator make the result not an integer? Am I right to think this has to do with the number of prime factors?
Because 12!/8!5! and 12!/9!5! are also integers, while 12!/10!5! isn't. If we had 11*2^3*3^2 left with 7! on the bottom then 8! would be 2^3 and 9! would be 3^2 both of which we can still reduce... while 10! gives us a factor of 5 which we don't have.
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Re: Which of the following is an integer? [#permalink]
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energetics wrote:
NickTW wrote:
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 only
E) I, II, and III


For III, I went through and crossed out everything that's reduced to visualize...
12*11*102*93*8*7*6*5*4*3*2*1 / 7*6*5*4*3*2*1 * 5*4*3*2*1

So you're left with 12*11*2*3 (from taking 5 and 3 from 10 and 9) = 792. Makes sense.

ScottTargetTestPrep
As a rule, you said m = n + p, then m!/(n!p!). When does the factorial in the denominator make the result not an integer? Am I right to think this has to do with the number of prime factors?
Because 12!/8!5! and 12!/9!5! are also integers, while 12!/10!5! isn't. If we had 11*2^3*3^2 left with 7! on the bottom then 8! would be 2^3 and 9! would be 3^2 both of which we can still reduce... while 10! gives us a factor of 5 which we don't have.


So you ask a really great qestion. So, as you've shown above, when m < n + p, there is no hard and fast rule to determien when m!/(n!p!) is an integer and when it is not.
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Re: Which of the following is an integer? [#permalink]
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This question gives you a great chance to review factorials, and once you answer this question you'll be well-prepared for other questions that involve factorials.

Here's an explanation on YouTube

https://www.youtube.com/watch?v=U7kymGI ... PfYZOmjn_0

Answer (E)
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Re: Which of the following is an integer? [#permalink]
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Re: Which of the following is an integer? [#permalink]
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