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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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New post 13 Mar 2016, 22:21
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  55% (hard)

Question Stats:

61% (02:01) correct 39% (02:08) wrong based on 74 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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Re: (4!)^n is a factor of 12!, but (4!)^(n+1) is not factor of 12!. What i  [#permalink]

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New post 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
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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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New post 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.
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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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New post 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.
Thus, the answer is C.


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

Answer: C


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

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New post 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.
Thus, the answer is C.


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
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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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New post 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

Answer: C


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]

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Re: (4!)^n is a factor of 12!, but (4!)^(n+1) is not factor of 12!. What i   [#permalink] 27 Mar 2019, 11:56
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