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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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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.
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Re: (4!)^n is a factor of 12!, but (4!)^(n+1) is not factor of 12!. What i
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14 Mar 2016, 06:30
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
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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. 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
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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. 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
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15 Mar 2016, 20:59
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
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15 Mar 2016, 21:05
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
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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. 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 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
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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
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?




Re: (4!)^n is a factor of 12!, but (4!)^(n+1) is not factor of 12!. What i &nbs
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