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

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Intern
Joined: 11 Mar 2015
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Re: Math Revolution and GMAT Club Contest! If the greatest common divisor  [#permalink]

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19 Dec 2015, 00:29
1
GCD =120

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

Hence n=7

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Joined: 05 Jun 2013
Posts: 35
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19 Dec 2015, 02:18
1
(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.

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19 Dec 2015, 02:18
1
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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20 Dec 2015, 02:19
1
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
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Re: Math Revolution and GMAT Club Contest! If the greatest common divisor  [#permalink]

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20 Dec 2015, 09:47
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:

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. _________________ Intern Joined: 09 Mar 2018 Posts: 6 Re: Math Revolution and GMAT Club Contest! If the greatest common divisor [#permalink] ### Show Tags 25 Apr 2018, 10:37 Prime factorize 120 = 2*2*2*3*5 Pick up the answer choices and fit those in the question. For e.g., picking 7 would give you - 9!, 5!, 11! All these three numbers will definitely contain at least 3 2's, one 3 and one 5. Hence D is the answer Manager Joined: 31 Jan 2018 Posts: 66 GMAT 1: 700 Q46 V40 Re: Math Revolution and GMAT Club Contest! If the greatest common divisor [#permalink] ### Show Tags 25 Apr 2018, 12:04 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: 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!

The answer must be 7 (D)

120 is basically a 5!

so all the numbers (n+2)!, (n-2) and (n+4)! should be divided completely by 5!

Looking into the options provided, only for n = 7 all the numbers are divided by 5!
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08 May 2019, 08:54
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Re: Math Revolution and GMAT Club Contest! If the greatest common divisor   [#permalink] 08 May 2019, 08:54

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