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If the greatest common divisor of (n-1)!, n!, and (n+1)! is 5040, what

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If the greatest common divisor of (n-1)!, n!, and (n+1)! is 5040, what  [#permalink]

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New post 14 Aug 2018, 16:47
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A
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E

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[Math Revolution GMAT math practice question]

If the greatest common divisor of (n-1)!, n!, and (n+1)! is 5040, what is the value of n?

A. 4
B. 5
C. 6
D. 7
E. 8

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Re: If the greatest common divisor of (n-1)!, n!, and (n+1)! is 5040, what  [#permalink]

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New post 14 Aug 2018, 17:10
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MathRevolution wrote:
[Math Revolution GMAT math practice question]

If the greatest common divisor of (n-1)!, n!, and (n+1)! is 5040, what is the value of n?

A. 4
B. 5
C. 6
D. 7
E. 8


Greatest common divisor of (n-2)!, n! and (n+1)! Will be (n-1)!..
So (n-1)!=5040=2*2*2*2*3*3*5*7=1*2*3*4*5*6*7=7!
Thus n-1=7.....n=8

E
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Re: If the greatest common divisor of (n-1)!, n!, and (n+1)! is 5040, what  [#permalink]

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New post 15 Aug 2018, 01:06
MathRevolution wrote:
[Math Revolution GMAT math practice question]

If the greatest common divisor of (n-1)!, n!, and (n+1)! is 5040, what is the value of n?

A. 4
B. 5
C. 6
D. 7
E. 8

\(5040 = 2^4 x 3^2 x 5^1 x 7^1\)

We also know, \(7! = 5040\)

\(GCF {(n-1)!, n!,\&\ (n+1)!) = 5040\)

Thus, the Minimum value of the group must have 7! ,

If, \((n - 1)! = 7!\) \(n = 8\), Thus Answer must be (E)
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Re: If the greatest common divisor of (n-1)!, n!, and (n+1)! is 5040, what  [#permalink]

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New post 15 Aug 2018, 06:47
=>

Since n! and (n+1)! are multiples of (n-1)!, (n-1)! is their gcd.
It follows that n – 1 = 7 or n = 8, since 7! = 5040.

Therefore, E is the answer.
Answer: E
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Re: If the greatest common divisor of (n-1)!, n!, and (n+1)! is 5040, what &nbs [#permalink] 15 Aug 2018, 06:47
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