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

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

by Max@Math Revolution » Thu Apr 21, 2016 12:30 am
If the greatest common divisor of (n+2)!, (n-1)!, and (n+4)! is 120, what is the value of n?

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


* A solution will be posted in two days.
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by 800_or_bust » Thu Apr 21, 2016 6:27 am
Max@Math Revolution wrote:If the greatest common divisor of (n+2)!, (n-1)!, and (n+4)! is 120, what is the value of n?

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


* A solution will be posted in two days.
I came up with (C) 6. My technique was to recognize that (n+2)! and (n+4)! would each share all of the factors of (n-1)!. So really the only thing we need to find is the lowest value of n such that (n-1)! is divisible by 120, because then we know that (n+2)! and (n+4)! are also divisible by that sum. And so that would be the greatest common divisor of all three.

Obviously, higher values of n would also share a common divisor of 120, but it wouldn't be the GREATEST common divisor.

With that said, testing values it's clear that the solution is n=6. At n=6, (n-1)! evaluates to be 5x4x3x2 = 120. And again we know with certainty we can at least factor out 120 from each of the other two factorials.
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by Brent@GMATPrepNow » Thu Apr 21, 2016 6:31 am
Max@Math Revolution wrote:If the greatest common divisor of (n+2)!, (n-1)!, and (n+4)! is 120, what is the value of n?

A. 4
B. 5
C. 6
D. 7
E. 3
Notice that:
1! = 1
2! = 2
3! = 6
4! = 24
5! = 120

So, the smallest factorial must be 5! for the GCD to be 120
The smallest of the three factorials is (n-1)!
So, we need (n-1)! to equal 5!
This means that n = 6

If n = 6, then (n+2)!, (n-1)!, and (n+4)! become 8!, 5! and 10!

Let's CONFIRM this conclusion
8! = (1)(2)(3)(4)(5)(6)(7)(8) = (120)(6)(7)(8)
5! = (1)(2)(3)(4)(5) = 120
10! = (1)(2)(3)(4)(5)(6)(7)(8)(9)(10) = (120)(6)(7)(8)(9)(10)

As you can see, 120 is a divisor of all three factorials.
In fact, 120 is the GREATEST COMMON divisor of all three factorials.

Answer: C

Cheers,
Brent
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by Max@Math Revolution » Wed Apr 27, 2016 12:47 am
If the greatest common divisor of (n+2)!, (n-1)!, and (n+4)! is 120, what is the value of n?

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


==>(n+2)!=(n+2)(n+1)n(n-1)!, (n-1)!,
and (n+4)!=(n+4)(n+3)(n+2)(n+1)n(n-1)!.
Then, the greatest common divisor becomes (n-1)!=120.
n-1=5 -> n=6.
Hence, the answer is C.
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