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

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If the greatest common divisor of (n+2)!, (n-2)!, and (n+4)! is 120,  [#permalink]

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13 Dec 2015, 02:54
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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! _________________ Most Helpful Community Reply Intern Joined: 06 Jul 2014 Posts: 15 WE: Supply Chain Management (Manufacturing) Re: If the greatest common divisor of (n+2)!, (n-2)!, and (n+4)! is 120, [#permalink] Show Tags 13 Dec 2015, 05:12 4 1 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 Solution- Since 5!=120 hence smallest of three numbers is 5. Hence n-2=5 or n=7 Ans-D General Discussion Intern Joined: 17 Aug 2014 Posts: 10 Re: If the greatest common divisor of (n+2)!, (n-2)!, and (n+4)! is 120, [#permalink] Show Tags 13 Dec 2015, 07:23 2 (n+2)!=(n+2)*(n+1)*n*(n-1)*(n-2)! (n-2)! (n+4)!=(n+4)*(n+3)*(n+2)*(n+1)*n*(n-1)*(n-2)! GCD=common factors=(n-2)! Given, GCD=120 n-2=5=>n=7 Ans D Intern Joined: 29 Sep 2014 Posts: 14 Re: If the greatest common divisor of (n+2)!, (n-2)!, and (n+4)! is 120, [#permalink] Show Tags 13 Dec 2015, 15:00 4 When considering the lowest common divisor of three factorials, we know that by the very nature of factorials, the (smallest number) factorial will be a divisor of any factorial of a larger number. This can be seen as $$(x+1)!=x!*(x+1)$$. If 120 is the largest common divisor of these three factorials then we know that the value of the smallest factorial must equal 120. If this were not the case then the largest common divisor would be a different number. The smallest factorial here is $$(n-2)!$$, and we know that $$5!=120$$, therefore $$n-2=5$$ and $$n=7$$. The answer is D. Intern Joined: 29 Aug 2013 Posts: 35 Location: Bangladesh GPA: 3.76 WE: Supply Chain Management (Transportation) Re: If the greatest common divisor of (n+2)!, (n-2)!, and (n+4)! is 120, [#permalink] Show Tags 14 Dec 2015, 04:12 1 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 Solution: (n+2)!=(n+2)(n+1)!=(n+2)(n+1)n!=(n+2)(n+1)n(n-1)!=(n+2)(n+1)n(n-1)(n-2)! (n-2)!=(n-2)! (n+4)!=(n+4)(n+3)!=(n+4)(n+3)(n+2)!=(n+4)(n+3)(n+2)(n+1)!=(n+4)(n+3)(n+2)(n+1)n!=(n+4)(n+3)(n+2)(n+1)n(n-1)!=(n+4)(n+3)(n+2)(n+1)n(n-1)(n-2)! hereby, Greatest common divisor= (n-2)! Given condition: (n-2)! will be 120 if we plug (n-2)! into answers, we get: A. 4 =(4-2)!=2!=2 B. 5 =(5-2)!=3!=6 C. 6 =(6-2)!=4!=24 D. 7 =(7-2)!=5!=120 E. 3=(3-2)!=1!=1 Answer: "D" Marshall & McDonough Moderator Joined: 13 Apr 2015 Posts: 1680 Location: India Re: If the greatest common divisor of (n+2)!, (n-2)!, and (n+4)! is 120, [#permalink] Show Tags 14 Dec 2015, 10:21 2 greatest common divisor of (n+2)!, (n-2)!, and (n+4)! is 120 Each of (n+2)!, (n-2)!, and (n+4)! has to be divisible by 120 120 = 5! (n+2)! has to be divisible by 5! --> n can be an integer >= 3 (n-2)! has to be divisible by 5! --> n can be an integer >= 7 (n+4)! has to be divisible by 5! --> n can be an integer >= 1 From the above values of n it can be seen that the minimum value of n has to be 7. Answer: D Manager Joined: 10 Aug 2015 Posts: 54 Concentration: General Management, Entrepreneurship GMAT 1: 730 Q48 V42 Re: If the greatest common divisor of (n+2)!, (n-2)!, and (n+4)! is 120, [#permalink] Show Tags 16 Dec 2015, 12:50 1 $$n$$ = _____________ ? Arrange this expression in order: $$( n - 2 ) !$$ , $$( n + 2 ) !$$ and $$( n + 4 ) !$$. Since these are factorials, it means that $$( n - 2 ) !$$ is contained within both $$( n + 2 )!$$ and $$( n + 4 ) !$$. So the greatest common factor (or divisor) of all these expressions is actually $$( n - 2 ) !$$. $$( n - 2 ) ! = 120$$ --> $$( n - 2 ) ! = 5 !$$ --> $$n = 7$$. The demonstration : $$( n - 2 ) ! = ( n - 2 ) * ( n - 3 ) * ( n - 4 ) * ( n - 5 ) * (....)$$. We don't know where this factorization stops, it depends on the value of n. But we don't need to know. $$( n + 2 ) ! = ( n + 2 ) * ( n + 1 ) * ( n ) * ( n - 1 ) * ( n - 2 ) !$$ $$( n + 4) ! = ( n + 4) * ( n + 3) * ( n + 2) * ( n + 1 ) * ( n ) * ( n - 1 ) * ( n - 2 ) !$$ So as you can see $$( n - 2 ) !$$ is actually the greatest common factor for all three expressions. You can be sure that it is the greatest common factor because none of the factors of ( n + 2 ) ! and ( n + 4) ! can be found in the factors of ( n - 2) ! . I hope that this is clear. Answer D. Intern Joined: 21 Jan 2013 Posts: 36 Concentration: General Management, Leadership Schools: LBS GPA: 3.82 WE: Engineering (Computer Software) Re: If the greatest common divisor of (n+2)!, (n-2)!, and (n+4)! is 120, [#permalink] Show Tags 18 Dec 2015, 23:06 1 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 -- Expanding : (n+2)! , (n-2)! and (n+4)! : (n+2)*(n+1)*(n)*(n-1)*(n-2)! (n-2)! (n+4)*(n+3)*(n+2)*(n+1)*(n)*(n-1)*(n-2)! Each of these factorials have a common factorial (n-2)! , so (n-2)! is one possible divisor. Given that a largest factor for any integer K is K itself, AND that we're looking for the GREATEST COMMON factor for the three factorials presented, it can be deduced that (n-2)! is indeed the greatest common factor looking for. any factor bigger than (n-2)! cannot be common across the three terms because it cannot divide (n-2)! - so cannot be common , and any factor of the three terms , lesser than (n-2)! cannot be the GCD because it cannot be greatest. Hence, (n-2)! is the GCD. Given : (n-2)! = 120 => n-2 = 5 Hence n = 7; Answer : D Math Expert Joined: 02 Sep 2009 Posts: 62637 Re: If the greatest common divisor of (n+2)!, (n-2)!, and (n+4)! is 120, [#permalink] Show Tags 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.
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Re: If the greatest common divisor of (n+2)!, (n-2)!, and (n+4)! is 120,  [#permalink]

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08 May 2019, 08:54
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Re: If the greatest common divisor of (n+2)!, (n-2)!, and (n+4)! is 120,   [#permalink] 08 May 2019, 08:54
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