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Which of the following is an integer?

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Which of the following is an integer?  [#permalink]

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New post Updated on: 07 Oct 2019, 02:36
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Which of the following is an integer?

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

Originally posted by NickTW on 18 Aug 2009, 20:23.
Last edited by Bunuel on 07 Oct 2019, 02:36, edited 2 times in total.
Renamed the topic, edited the question and added the OA.
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Re: Which of the following is an integer?  [#permalink]

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New post 08 May 2015, 16:47
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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.

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Re: Which of the following is an integer?  [#permalink]

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New post 18 Aug 2009, 20:30
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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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New post 18 Aug 2009, 20:45
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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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New post 19 Aug 2009, 09:32
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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]

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New post 03 May 2015, 02:42
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]

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New post 03 Dec 2017, 18:39
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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


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
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Re: Which of the following is an integer?  [#permalink]

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New post 20 Jan 2019, 09:00
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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New post 22 Jan 2019, 18:19
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 is integer?  [#permalink]

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Re: Which is integer?   [#permalink] 07 Oct 2019, 02:36
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