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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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(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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(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.
[Reveal] Spoiler: OA

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

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14 Mar 2016, 06:30
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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

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Math Revolution GMAT Instructor
Joined: 16 Aug 2015
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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, 18:46
(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.
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Kudos [?]: 3035 [0], given: 0

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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.

Kudos [?]: 1169 [0], given: 888

Math Expert
Joined: 02 Aug 2009
Posts: 5201

Kudos [?]: 5826 [1], given: 117

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
KUDOS
Expert's post
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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Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
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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
KUDOS
Expert's post
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

Kudos [?]: 5826 [1], given: 117

SC Moderator
Joined: 13 Apr 2015
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Kudos [?]: 1169 [0], given: 888

Location: India
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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!!

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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, 07:51
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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?

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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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