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Math Revolution and GMAT Club Contest! If the greatest common divisor

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Re: Math Revolution and GMAT Club Contest! If the greatest common divisor [#permalink]

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New post 19 Dec 2015, 01:29
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GCD =120

120 = 5!
So (n-2)=5!

Hence n=7

Answer choice is D !!

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Re: Math Revolution and GMAT Club Contest! If the greatest common divisor [#permalink]

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New post 19 Dec 2015, 03:18
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(n+2)!, (n-2)!, and (n+4)! is 120
GCD of two numbers would be the lowest powers of prime numbers from each of the two numbers!

We can try out one of answer choices say 3 n=3, which is GCD of 5!, 3!,9!
5! = 1*2*3*2*2*5 , 3! = 1*2*3, 9!= 1*2*3*2*2*5*2*3*7*2*2*2*3*3
Therefore of GCD when n =3 is 6.

Since all 3 numbers are factorials , obviously the middle number is factor of other two.
Therefore to get GCD of 120, (n-2)! = 120 which is (n-2)! = 5! and therefore n =7.

Hence answer choice D

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Re: Math Revolution and GMAT Club Contest! If the greatest common divisor [#permalink]

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New post 19 Dec 2015, 03:18
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We can re-write the factors to see what is there in common:

(n+4)! = (n+4)!*(n+3)!*(n+2)!*(n+1)!*n*(n-1)!*(n-2)!
(n+2)! = (n+2)!*(n+1)!*n*(n-1)!*(n-2)!
(n-2)! = (n-2)!

greatest common divisor of (n+2)!, (n-2)!, and (n+4)! is (n-2)! because this common in both. it can divide all the three numbers. its value is given as 120.

hence (n-2)! = 120 = 5*4*3*2*1 = 5! = (7-2)!

n=7.

Option D is the correct answer.

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Re: Math Revolution and GMAT Club Contest! If the greatest common divisor [#permalink]

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New post 20 Dec 2015, 03:19
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GCD depends on least of (n+2)!, (n-2)! and (n+2)!,
(n-2)! being the least of the three it shall be a multiple of 120
n=7 gives 5!=120
Answer is D

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Re: Math Revolution and GMAT Club Contest! If the greatest common divisor [#permalink]

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New post 20 Dec 2015, 10:47
Bunuel wrote:

Math Revolution and GMAT Club Contest Starts!



QUESTION #11:

If the greatest common divisor of (n+2)!, (n-2)!, and (n+4)! is 120, what is the value of n?

A. 4
B. 5
C. 6
D. 7
E. 3


Check conditions below:


[Reveal] Spoiler:

Math Revolution and GMAT Club Contest

The Contest Starts November 28th in Quant Forum


We are happy to announce a Math Revolution and GMAT Club Contest

For the following four (!) weekends we'll be publishing 4 FRESH math questions per weekend (2 on Saturday and 2 on Sunday).

To participate, you will have to reply with your best answer/solution to the new questions that will be posted on Saturday and Sunday at 9 AM Pacific.
Then a week later, the forum moderator will be selecting 2 winners who provided most correct answers to the questions, along with best solutions. Those winners will get 6-months access to GMAT Club Tests.

PLUS! Based on the answers and solutions for all the questions published during the project ONE user will be awarded with ONE Grand prize:

PS + DS course with 502 videos that is worth $299!



All announcements and winnings are final and no whining :-) GMAT Club reserves the rights to modify the terms of this offer at any time.


NOTE: Test Prep Experts and Tutors are asked not to participate. We would like to have the members maximize their learning and problem solving process.

Thank you!



MATH REVOLUTION OFFICIAL SOLUTION:

\((n+2)!=(n+2)(n+1)n(n-1)(n-2)!\) and \((n+4)!=(n+4)(n+3)(n+2)(n+1)n(n-1)(n-2)!\).

From this, we can observe that (n+4)!, (n+2)! and (n-2)! include (n-2)!, which means the greatest common divisor for them is (n-2)!. So, since we get n-2=5 and n=7 from \((n-2)!=120=5!\), the correct answer is D.
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Re: Math Revolution and GMAT Club Contest! If the greatest common divisor   [#permalink] 01 Sep 2017, 11:41

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