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# (4!)^n is a factor of 12!, but (4!)^(n+1) is not factor of 12!. What i

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Math Revolution GMAT Instructor
Joined: 16 Aug 2015
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(4!)^n is a factor of 12!, but (4!)^(n+1) is not factor of 12!. What i [#permalink]

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13 Mar 2016, 22:21
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Difficulty:

45% (medium)

Question Stats:

67% (01:09) correct 33% (01:25) wrong based on 76 sessions

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(4!)^n is a factor of 12!, but (4!)^(n+1) is not factor of 12!. What is the value of n?

A. 1
B. 2
C. 3
D. 4
E. 5

* A solution will be posted in two days.

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MathRevolution: Finish GMAT Quant Section with 10 minutes to spare
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"Only $79 for 3 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" SC Moderator Joined: 13 Apr 2015 Posts: 1684 Location: India Concentration: Strategy, General Management GMAT 1: 200 Q1 V1 GPA: 4 WE: Analyst (Retail) Re: (4!)^n is a factor of 12!, but (4!)^(n+1) is not factor of 12!. What i [#permalink] ### Show Tags 14 Mar 2016, 06:30 1 12! = 12 * 11 * 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 4! = 24 Question can be rephrased as how many 24's are there in 12! 12 * 2 = 24 4 * 6 = 24 3 * 8 = 24 Value of n = 3 Answer: C Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 5562 GMAT 1: 800 Q59 V59 GPA: 3.82 Re: (4!)^n is a factor of 12!, but (4!)^(n+1) is not factor of 12!. What i [#permalink] ### Show Tags 15 Mar 2016, 18:46 (4!)^n is a factor of 12!, but (4!)^(n+1) is not factor of 12!. What is the value of n? A. 1 B. 2 C. 3 D. 4 E. 5 -> 12!=210(integers) is derived. (4!)n=(23n)(3n) and in order to satisfy (4!)n+1=(23n+3)(3n+1), n should be 3. Thus, the answer is C. _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only$79 for 3 month Online Course"
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SC Moderator
Joined: 13 Apr 2015
Posts: 1684
Location: India
Concentration: Strategy, General Management
GMAT 1: 200 Q1 V1
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WE: Analyst (Retail)
Re: (4!)^n is a factor of 12!, but (4!)^(n+1) is not factor of 12!. What i [#permalink]

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15 Mar 2016, 18:52
MathRevolution wrote:
(4!)^n is a factor of 12!, but (4!)^(n+1) is not factor of 12!. What is the value of n?

A. 1
B. 2
C. 3
D. 4
E. 5

-> 12!=210(integers) is derived. (4!)n=(23n)(3n) and in order to satisfy (4!)n+1=(23n+3)(3n+1), n should be 3.

Can you please explain the approach taken? I am not able to understand the solution in its current form.
Math Expert
Joined: 02 Aug 2009
Posts: 5875
Re: (4!)^n is a factor of 12!, but (4!)^(n+1) is not factor of 12!. What i [#permalink]

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15 Mar 2016, 20:59
1
Vyshak wrote:
12! = 12 * 11 * 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1

4! = 24

Question can be rephrased as how many 24's are there in 12!

12 * 2 = 24
4 * 6 = 24
3 * 8 = 24

Value of n = 3

hi Vyshak,

another way to find 24 in 12! is

Now 24= 2^3 *3..
so we have to find how many 2^3 in 12!..
lets find 2s first= 12/2 + 12/2^2 +12/2^3 +12/2^4= 6+3+1+ less than 1= 10..
since 8=2^3 divide 10 by 3= 3.33..
we have to take integer value again.. so there are THREE 8s and thus THREE 24s in 12!.. so n=3

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Math Expert
Joined: 02 Aug 2009
Posts: 5875
Re: (4!)^n is a factor of 12!, but (4!)^(n+1) is not factor of 12!. What i [#permalink]

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15 Mar 2016, 21:05
1
Vyshak wrote:
MathRevolution wrote:
(4!)^n is a factor of 12!, but (4!)^(n+1) is not factor of 12!. What is the value of n?

A. 1
B. 2
C. 3
D. 4
E. 5

-> 12!=210(integers) is derived. (4!)n=(23n)(3n) and in order to satisfy (4!)n+1=(23n+3)(3n+1), n should be 3.

Can you please explain the approach taken? I am not able to understand the solution in its current form.

Hi,

It took me 20 seconds to grasp what was happening, but I think the solution here means-
$$12!=2^{10}$$(integers) is derived.
$$(4!)^n=(2^{3n})(3^n)$$ and in order to satisfy $$(4!)^{n+1}=(2^{3n+3})(3^{n+1})$$, n should b 3

Hope it is better now
_________________

Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html

GMAT online Tutor

SC Moderator
Joined: 13 Apr 2015
Posts: 1684
Location: India
Concentration: Strategy, General Management
GMAT 1: 200 Q1 V1
GPA: 4
WE: Analyst (Retail)
Re: (4!)^n is a factor of 12!, but (4!)^(n+1) is not factor of 12!. What i [#permalink]

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15 Mar 2016, 21:09
chetan2u wrote:
Vyshak wrote:
MathRevolution wrote:
(4!)^n is a factor of 12!, but (4!)^(n+1) is not factor of 12!. What is the value of n?

A. 1
B. 2
C. 3
D. 4
E. 5

-> 12!=210(integers) is derived. (4!)n=(23n)(3n) and in order to satisfy (4!)n+1=(23n+3)(3n+1), n should be 3.

Can you please explain the approach taken? I am not able to understand the solution in its current form.

Hi,

It took me 20 seconds to grasp what was happening, but I think the solution here means-
$$12!=2^{10}$$(integers) is derived.
$$(4!)^n=(2^{3n})(3^n)$$ and in order to satisfy $$(4!)^{n+1}=(2^{3n+3})(3^{n+1})$$, n should b 3

Hope it is better now

Yup. I get it now!!
Intern
Joined: 31 Jul 2017
Posts: 7
Re: (4!)^n is a factor of 12!, but (4!)^(n+1) is not factor of 12!. What i [#permalink]

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24 Sep 2017, 11:30
Vyshak wrote:
12! = 12 * 11 * 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1

4! = 24

Question can be rephrased as how many 24's are there in 12!

12 * 2 = 24
4 * 6 = 24
3 * 8 = 24

Value of n = 3

What made you think the answer is 3. Like why couldn't it be 4 or 12? I understand 12 isn't in the answer, but both 3 and 4 is. Other than brute force, is there a way to identify this?
Re: (4!)^n is a factor of 12!, but (4!)^(n+1) is not factor of 12!. What i   [#permalink] 24 Sep 2017, 11:30
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