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If the greatest common divisor of (n + 2)!, (n - 1)!, and (n + 4)! is

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Bunuel
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If the greatest common divisor of (n + 2)!, (n - 1)!, and (n + 4)! is [#permalink] New post   08 Dec 2022, 02:30
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If the greatest common divisor of (n + 2)!, (n - 1)!, and (n + 4)! is 120, what is the value of the positive integer n?

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

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D
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chetan2u
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Re: If the greatest common divisor of (n + 2)!, (n - 1)!, and (n + 4)! is [#permalink] New post   08 Dec 2022, 08:10
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Bunuel wrote:
If the greatest common divisor of (n + 2)!, (n - 1)!, and (n + 4)! is 120, what is the value of the positive integer n?

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


Greatest common divisor of (n + 2)!, (n - 1)!, and (n + 4)! is the smallest of all three, that is (n-1)!.

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

Therefore, (n-1)!=120=5*4*3*2*1=5!.
Or, (n-1)=5……n=6


D
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Re: If the greatest common divisor of (n + 2)!, (n - 1)!, and (n + 4)! is [#permalink] New post   08 Dec 2022, 10:06
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Look for the greatest common factors in (n + 2)!, (n - 1)!, and (n + 4)! --> definitely it is (n-1)!

(n-1)! = 120 = 5!
-> n-1 = 5
-> n = 6

hence D is the answer!
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Re: If the greatest common divisor of (n + 2)!, (n - 1)!, and (n + 4)! is [#permalink] New post   08 Dec 2022, 13:26
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Bunuel wrote:
If the greatest common divisor of (n + 2)!, (n - 1)!, and (n + 4)! is 120, what is the value of the positive integer n?

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


120 = 5 * 4 * 3 * 2 = 5!

The GCD of (n + 2)!, (n - 1)!, and (n + 4)! is (n-1) !

(n - 1)! = 5!

n = 6

Option D
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Re: If the greatest common divisor of (n + 2)!, (n - 1)!, and (n + 4)! is [#permalink]
08 Dec 2022, 13:26
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