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If the greatest common divisor of (n+2)!, (n2)!, and (n+4)! is 120,
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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)!, (n2)!, 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 ContestThe 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 6months 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.
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Re: If the greatest common divisor of (n+2)!, (n2)!, and (n+4)! is 120,
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13 Dec 2015, 05:12
If the greatest common divisor of (n+2)!, (n2)!, 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 n2=5 or n=7 AnsD




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Re: If the greatest common divisor of (n+2)!, (n2)!, and (n+4)! is 120,
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13 Dec 2015, 07:23
(n+2)!=(n+2)*(n+1)*n*(n1)*(n2)! (n2)! (n+4)!=(n+4)*(n+3)*(n+2)*(n+1)*n*(n1)*(n2)!
GCD=common factors=(n2)! Given, GCD=120 n2=5=>n=7 Ans D



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Re: If the greatest common divisor of (n+2)!, (n2)!, and (n+4)! is 120,
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13 Dec 2015, 15:00
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 \((n2)!\), and we know that \(5!=120\), therefore \(n2=5\) and \(n=7\).
The answer is D.



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Re: If the greatest common divisor of (n+2)!, (n2)!, and (n+4)! is 120,
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14 Dec 2015, 04:12
If the greatest common divisor of (n+2)!, (n2)!, 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(n1)!=(n+2)(n+1)n(n1)(n2)! (n2)!=(n2)! (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(n1)!=(n+4)(n+3)(n+2)(n+1)n(n1)(n2)!
hereby, Greatest common divisor= (n2)!
Given condition: (n2)! will be 120
if we plug (n2)! into answers, we get:
A. 4 =(42)!=2!=2 B. 5 =(52)!=3!=6 C. 6 =(62)!=4!=24 D. 7 =(72)!=5!=120 E. 3=(32)!=1!=1
Answer: "D"



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Re: If the greatest common divisor of (n+2)!, (n2)!, and (n+4)! is 120,
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14 Dec 2015, 10:21
greatest common divisor of (n+2)!, (n2)!, and (n+4)! is 120
Each of (n+2)!, (n2)!, 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 (n2)! 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



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Re: If the greatest common divisor of (n+2)!, (n2)!, and (n+4)! is 120,
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16 Dec 2015, 12:50
\(n\) = _____________ ?
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.



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Re: If the greatest common divisor of (n+2)!, (n2)!, and (n+4)! is 120,
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18 Dec 2015, 23:06
QUESTION #11:
If the greatest common divisor of (n+2)!, (n2)!, and (n+4)! is 120, what is the value of n?
A. 4 B. 5 C. 6 D. 7 E. 3
 Expanding : (n+2)! , (n2)! and (n+4)! :
(n+2)*(n+1)*(n)*(n1)*(n2)! (n2)! (n+4)*(n+3)*(n+2)*(n+1)*(n)*(n1)*(n2)!
Each of these factorials have a common factorial (n2)! , so (n2)! 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 (n2)! is indeed the greatest common factor looking for. any factor bigger than (n2)! cannot be common across the three terms because it cannot divide (n2)!  so cannot be common , and any factor of the three terms , lesser than (n2)! cannot be the GCD because it cannot be greatest.
Hence, (n2)! is the GCD.
Given : (n2)! = 120 => n2 = 5 Hence n = 7;
Answer : D



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Re: If the greatest common divisor of (n+2)!, (n2)!, and (n+4)! is 120,
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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)!, (n2)!, 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 ContestThe 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 6months 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:\((n+2)!=(n+2)(n+1)n(n1)(n2)!\) and \((n+4)!=(n+4)(n+3)(n+2)(n+1)n(n1)(n2)!\). From this, we can observe that (n+4)!, (n+2)! and (n2)! include (n2)!, which means the greatest common divisor for them is (n2)!. So, since we get n2=5 and n=7 from \((n2)!=120=5!\), the correct answer is D.
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Re: If the greatest common divisor of (n+2)!, (n2)!, and (n+4)! is 120,
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Re: If the greatest common divisor of (n+2)!, (n2)!, and (n+4)! is 120,
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