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# 12!/(3^4*5!*2^6)=

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Math Expert
Joined: 02 Sep 2009
Posts: 58381

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11 Jun 2015, 04:40
2
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Difficulty:

15% (low)

Question Stats:

82% (01:32) correct 18% (02:09) wrong based on 136 sessions

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$$\frac{12!}{(3^4*5!*2^6)}=$$

(A) 2,210
(B) 770
(C) 480
(D) 77
(E) 35

Kudos for a correct solution.

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Math Expert
Joined: 02 Aug 2009
Posts: 7959

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12 Jun 2015, 01:33
7
1
Bunuel wrote:
$$\frac{12!}{(3^4*5!*2^6)}=$$

(A) 2,210
(B) 770
(C) 480
(D) 77
(E) 35

Kudos for a correct solution.

hello all
there is no requirement to carry out lengthy calculations..
since we are racing against time , we should try and use the properties of the numbers we see in front..

here we can easily see numerator 12! consist of prime numbers 7 and 11, which are not being cancelled out by denominator..
so ans has to be a multiple of 77... only B and D are left ..

now 12! also consists of two*5s(one each in 5 and 10).. however the denominator has only one 5 in 5!..
the ans has to be multiple of 77*5.. only B is left ..

ans B 770
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11 Jun 2015, 05:03
1
Bunuel wrote:
$$\frac{12!}{(3^4*5!*2^6)}=$$

(A) 2,210
(B) 770
(C) 480
(D) 77
(E) 35

Kudos for a correct solution.

$$\frac{12!}{(3^4*5!*2^6)}=\frac{(12*11*10*9*8*7*6*5!)}{(3^4*5!*2^6)}$$

i.e. $$\frac{12!}{(3^4*5!*2^6)}=\frac{(12*11*10*9*8*7*6)}{(3^4*2^6)}$$

i.e. $$\frac{12!}{(3^4*5!*2^6)}=\frac{(2^2*3*11*2*5*3^2*2^3*7*2*3)}{(3^4*2^6)}$$

i.e. $$\frac{12!}{(3^4*5!*2^6)}=\frac{(2^7*3^4*11*5*7)}{(3^4*2^6)}$$

i.e. $$\frac{12!}{(3^4*5!*2^6)}=\frac{(2*11*5*7)}{(1)}$$

i.e. $$\frac{12!}{(3^4*5!*2^6)}=(770)$$

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Updated on: 12 Jun 2015, 02:12
1
Bunuel wrote:
$$\frac{12!}{(3^4*5!*2^6)}=$$

(A) 2,210
(B) 770
(C) 480
(D) 77
(E) 35

Kudos for a correct solution.

$$\frac{12!}{(3^4*5!*2^6)}=$$

$$=\frac{12*11*10*9*8*7*6*5*4*3*2}{(3*3*3*3)*(5*4*3*2*1)*(2*2*2*2*2*2)}=$$

$$=\frac{(2*2*3)*11*(2*5)*(3*3)*(2*2*2)*7*(2*3)*5*(2*2)*3*2}{(3*3*3*3)*(5*(2*2)*3*2)*(2*2*2*2*2*2)}=$$

$$=\frac{2^{10}*3^5*11*5^2*7}{3^5*5*2^9}=$$

$$=2*11*5*7=2*11*35=2*385=770=B$$

Originally posted by bluesquare on 12 Jun 2015, 01:18.
Last edited by Bunuel on 12 Jun 2015, 02:12, edited 2 times in total.
Formatting.
Math Expert
Joined: 02 Sep 2009
Posts: 58381

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15 Jun 2015, 02:28
Bunuel wrote:
$$\frac{12!}{(3^4*5!*2^6)}=$$

(A) 2,210
(B) 770
(C) 480
(D) 77
(E) 35

Kudos for a correct solution.

MANHATTAN GMAT OFFICIAL SOLUTION:

Since 5! and 12! share the common terms (5)(4)(3)(2)(1), these terms can be canceled from both the numerator and denominator. Thus, we can rewrite $$\frac{12!}{(3^4*5!*2^6)}$$ as:

$$\frac{12*10*9*8*7*6}{(3^4*2^6)}=$$

We need to cancel the 2's and 3's in the denominator, so we identify these common factors in the numerator.

$$\frac{(2^2*3)*11*(2*5)*(3^2)*(2^3)*7*(2*3)}{(3^4*2^6)}$$

Group the common factors of 2 and 3 in the numerator, then cancel and compute:

$$\frac{(2^7*3^4)*11*5*7}{(3^4*2^6)}=2*11*5*7=770$$

Note: If you do not have time to do all the canceling involved in this problem, you can at least rule out answer choices C (480) and E (35), because the correct answer must be a multiple of 11. The 12! in the numerator is a multiple of 11, and the denominator contains no factors of 11 that would cancel this term. You could use the same argument to eliminate non-multiples of 7; however, the divisibility rule for 7 is complicated, and on typical GMAT problems, testing by long division or inspection will be just as fast. In any event, you could eliminate answer choices C (480) and A (2,210) in this manner.
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29 Oct 2018, 04:55
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Re: 12!/(3^4*5!*2^6)=   [#permalink] 29 Oct 2018, 04:55
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