Which of the following is an integer?

Question

I.  \frac{12!}{6!}

II.  \frac{12!}{8!}

III.  \frac{12!}{7!5!}

(A) I only
(B) II only
(C) III only
(D) I and II only
(E) I, II, and III

ScottTargetTestPrep · Accepted Answer

Before actually solving this problem, let's review how factorials can be expanded and expressed. As as example, we can use 5!.
 
5! could be expressed as:
 
5! 
 
5 x 4! 
 
5 x 4 x 3! 
 
5 x 4 x 3 x 2! 
 
5 x 4 x 3 x 2 x 1!
 
Understanding how this factorial expansion works will help us work our way through each answer choice, especially answer choices 1 and 2.
 
I. 12!/6! 
 
Since we know that factorials can be expanded, we now know that:
 
12! = 12 x 11 x 10 x 9 x 8 x 7 x 6!
 
Plugging this in for answer choice 1, we have:
 
(12 x 11 x 10 x 9 x 8 x 7 x 6!)/6! = 12 x 11 x 10 x 9 x 8 x 7, which is an integer.
 
II. 12!/8! 
 
Once again, since we know that factorials can be expanded, we now know that:
 
12! = 12 x 11 x 10 x 9 x 8!
 
Plugging this in for answer choice 2, we have:
 
(12 x 11 x 10 x 9 x 8!)/8! = 12 x 11 x 10 x 9, which is an integer.
 
III. 12!/(7!5!) 
 
Once again, since we know that factorials can be expanded, we now know that:
 
12! = 12 x 11 x 10 x 9 x 8 x 7!
 
Plugging this in for answer choice 3 gives us:
 
(12 x 11 x 10 x 9 x 8 x 7!)/(7!5!) 
 
(12 x 11 x 10 x 9 x 8)/(5 x 4 x 3 x 2 x 1)
 
(12 x 11 x 10 x 9 x 8)/(12 x 10 x 1)
 
11 x 9 x 8, which is an integer.
 
We see that the quantities in Roman numerals I, II and III are all integers.

Alternate solution: 
 
For any positive integers m, n and p,
 
1) If m > n, then m!/n! is always an integer.
 
2) If m = n + p, then m!/(n!p!) is always an integer (which is in fact mCp).
 
From the above two facts, we see that all three quotients in the Roman numerals must be integers.

Answer: E

EMPOWERgmatRichC · Answer

Hi All,

Beyond the arithmetic that sdrandom1 pointed out, these calculations fall into some patterns that you are likely to see (in some form) on Test Day.

12!/6!

When dealing with INDIVIDUAL factorials, if the factorial in the numerator is GREATER than the factorial in the denominator, then you will end up with an integer. Here, since 12 > 6, 12!/6! will simplify to an integer.

12!/7!5!

You might not have studied the Combination Formula yet, but this calculation is what you would end up with when answering the question "how many different combinations of 7 people can you form from a group of 12 people?" In these types of "Combination" calculations, the number of groups will always be an integer, so 12!/7!5! will simplify to an integer.

Final Answer:  E

GMAT assassins aren't born, they're made, 
Rich

alwynjoseph · Answer

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

sdrandom1 · Answer

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.

IanStewart · Answer

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.

thangvietname · Answer

case 1 and case 2 are simple but case 3 is harder. for case 3, we have to write down all the factors.

energetics · Answer

For III, I went through and crossed out everything that's reduced to visualize...
12*11* 10 2* 9 3* 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.

ScottTargetTestPrep · Answer

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.

CrushTheQuant · Answer

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)

bumpbot · Answer

Automated notice from GMAT Club BumpBot: 

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

 This post was generated automatically.

Which of the following is an integer?

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

Which of the following is an integer?

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards