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The function F(n) is defined as the product of all the conse

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The function F(n) is defined as the product of all the conse [#permalink]

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15 Oct 2012, 10:16
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Manhattan weekly challenge problem

The function F(n) is defined as the product of all the consecutive positive integers between 1 and n^2, inclusive, whereas the function G(n) is defined as the product of the squares of all the consecutive positive integers between 1 and n, inclusive. The exponent on 2 in the prime factorization of F(3)/G(3) is

(A) 1
(B) 2
(C) 3
(D) 4
(E) 5
[Reveal] Spoiler: OA

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Re: The function F(n) is defined as the product of all the conse [#permalink]

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15 Oct 2012, 10:29
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thevenus wrote:
Manhattan weekly challenge problem

The function F(n) is defined as the product of all the consecutive positive integers between 1 and n^2, inclusive, whereas the function G(n) is defined as the product of the squares of all the consecutive positive integers between 1 and n, inclusive. The exponent on 2 in the prime factorization of F(3)/G(3) is

(A) 1
(B) 2
(C) 3
(D) 4
(E) 5

The function F(n) is defined as the product of all the consecutive positive integers between 1 and n^2, inclusive, thus $$F(3)=1*2*3*...*9=9!$$.

The function G(n) is defined as the product of the squares of all the consecutive positive integers between 1 and n, inclusive, thus $$G(3)=1^2*2^2*3^2=3!*3!$$.

$$\frac{F(3)}{G(3)}=\frac{9!}{3!*3!}=\frac{4*5*6*7*8*9}{6}=2^2*5*7*2^3*9=2^5*(5*7*9)$$.

The power of 2 is 5.

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Re: The function F(n) is defined as the product of all the conse [#permalink]

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15 Oct 2012, 10:30
F(3)/G(3)
=product(1 to 3 ^2) / 1.2^2.3^2
=1.2.3.4.5.6.7.8.9/1.4.9
=1.2.3.(2^2).5.(2.3).7.(2^3).9/1.(2^2).9
=1.(2^7).3.5.7.9/1.(2^2).9
Loof for 2^7/2^2=2^5 ----Exponent 5

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Re: The function F(n) is defined as the product of all the conse [#permalink]

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08 Sep 2013, 05:25
Bumping for review and further discussion.
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Re: The function F(n) is defined as the product of all the conse [#permalink]

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23 Sep 2014, 14:08
Hello from the GMAT Club BumpBot!

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Re: The function F(n) is defined as the product of all the conse [#permalink]

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23 Sep 2014, 20:22
Did not conceive the problem which is just to find the degree of 2 in final. So

9! factorization is

9=3*3 (remove)
8=2*2*2
7=7
6=3*2
5=5
4=2*2 (remove)
3=3
2=2

36 factorization is
3, 4, 3 (remove)
Five 2s are bolded

it is E
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Re: The function F(n) is defined as the product of all the conse [#permalink]

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23 Sep 2014, 23:59
This problem takes more time to read than to solve

$$F(n) = (n^2)!$$

$$F(3) = 9!$$ [Powers of 2 = 7] ............. (1)

$$G(n) = 1^2 * 2^2 * 3^2 .......... n^2 = (n!)^2$$

$$G(3) = (3!)^2 = 36$$ [Powers of 2 = 2] ............ (2)

$$\frac{(1)}{(2)}$$ [Powers of 2 = 7-2 = 5]

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Re: The function F(n) is defined as the product of all the conse [#permalink]

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21 May 2016, 12:11
Hello from the GMAT Club BumpBot!

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Re: The function F(n) is defined as the product of all the conse   [#permalink] 21 May 2016, 12:11
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