I'm going with [spoiler]n=3.
[/spoiler]
12! factored is (2^10)(3^5)(5^2)(7)(11)
4! factored is (2^3)(3)
Therefore, (4!)^3 is (2^9)(3^3) which is contained within the factors of 12!
and (4!)^4 is (2^12)(3^4) which is not contained within the factors of 12!
PS
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Solution:
Note that 12! should have (4!)^n as factor and not have (4!)^(n+1) as factor.
12! = (2^10)*(3^5)*(5^2)*7*11.
4! = (2^3)*3.
So (4!)^n = (2^3n)*(3^n).
Also (4!)^(n+1) = 2^(3n+3) * 3^(n+1)
Since (4!)^n is dividing 12!, 3n <= 10 and n<=5.
Or n<=(10/3 = 3 1/3) and n<=5.
Or n <= 3 1/3 and n <= 5.
Since (4!)^(n+1) is not dividing 12!, 3n+3 > 10 or n+1 > 5.
Or n > 7/3 or n > 4.
Or n > 2 1/3 or n >4.
Combining we have 2 1/3 < n < = 3 1/3.
The only possible value is n = 3.
Note that 12! should have (4!)^n as factor and not have (4!)^(n+1) as factor.
12! = (2^10)*(3^5)*(5^2)*7*11.
4! = (2^3)*3.
So (4!)^n = (2^3n)*(3^n).
Also (4!)^(n+1) = 2^(3n+3) * 3^(n+1)
Since (4!)^n is dividing 12!, 3n <= 10 and n<=5.
Or n<=(10/3 = 3 1/3) and n<=5.
Or n <= 3 1/3 and n <= 5.
Since (4!)^(n+1) is not dividing 12!, 3n+3 > 10 or n+1 > 5.
Or n > 7/3 or n > 4.
Or n > 2 1/3 or n >4.
Combining we have 2 1/3 < n < = 3 1/3.
The only possible value is n = 3.
Rahul Lakhani
Quant Expert
Gurome, Inc.
https://www.GuroMe.com
On MBA sabbatical (at ISB) for 2011-12 - will stay active as time permits
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Quant Expert
Gurome, Inc.
https://www.GuroMe.com
On MBA sabbatical (at ISB) for 2011-12 - will stay active as time permits
1-800-566-4043 (USA)
+91-99201 32411 (India)
