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# If the greatest common divisor of (n+2)!, (n-1)!, and (n+4)! is 120, w

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If the greatest common divisor of (n+2)!, (n-1)!, and (n+4)! is 120, w  [#permalink]

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21 Apr 2016, 00:30
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Difficulty:

25% (medium)

Question Stats:

74% (01:51) correct 26% (02:11) wrong based on 106 sessions

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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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Re: If the greatest common divisor of (n+2)!, (n-1)!, and (n+4)! is 120, w  [#permalink]

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21 Apr 2016, 00:46
1
(n + 4)! > (n + 2)! > (n - 1)! --> GCD of the 3 numbers = (n - 1)!

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

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Re: If the greatest common divisor of (n+2)!, (n-1)!, and (n+4)! is 120, w  [#permalink]

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21 Apr 2016, 06:32
1
MathRevolution 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.

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.

Cheers,
Brent
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Re: If the greatest common divisor of (n+2)!, (n-1)!, and (n+4)! is 120, w  [#permalink]

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22 Apr 2016, 10:42
1
MathRevolution 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.

120 = 5 x 4 x 3 x 2 x 1

No try to substitute the options in place of n in the form

The problem states that 120 is the greatest common divisor of (n+2)!, (n-1)!, and (n+4)! , so the three factorials must be greater than 120 ( Rather 5! )

Now find out -

A. 4

(n+2)!, (n-1)!, and (n+4)! => 6! , 3! , 8!

B. 5

(n+2)!, (n-1)!, and (n+4)! => 7! , 4! , 9!

C. 6

(n+2)!, (n-1)!, and (n+4)! => 8! ,5! , 10!

D. 7

(n+2)!, (n-1)!, and (n+4)! => 9! , 8! , 11!

E. 3

(n+2)!, (n-1)!, and (n+4)! => 5! , 2! , 7!

Now check we have 2 contenders (C) and (D)

While we are confident about (C) lets check why not (D)

9! = 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1
8! = 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1
11! = 11 x 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1

Now I think you got it , yes the least common factor of 9! , 8! , 11! is 8! not 5!

Phew , lengthy post HUH...
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Re: If the greatest common divisor of (n+2)!, (n-1)!, and (n+4)! is 120, w  [#permalink]

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27 Apr 2016, 00:47
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.
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Re: If the greatest common divisor of (n+2)!, (n-1)!, and (n+4)! is 120, w  [#permalink]

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27 May 2017, 03:37
2
Well , the case with factorials is that:

Factorial of any positive integer is Always a factor of its successive factorials (eg, 4! will be a factor of 5!, 6!, 7!, 8!,... and so on)

And Factorial of any positive integer is Always a multiple of previous factorials (eg, 6! is a multiple of 5!, 4!, 3!, ... and so on)

So, on the basis of this information we can say that if we are given various factorials, the smallest of them will be the GCD/HCF of the lot and the largest of them will be the LCM of the lot.
Eg - if we are given 7!, 62!, 35! and 19! - then 7! will be their GCD and 62! will be their LCM.

Now, we are given (n+2)!, (n-1)! and (n+4)!. Their GCD will be the smallest one, i.e., (n-1)!

So, (n-1)! = 120, which is 5!
Thus n-1 = 5 or n = 6.

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Re: If the greatest common divisor of (n+2)!, (n-1)!, and (n+4)! is 120, w  [#permalink]

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04 Feb 2019, 07:30
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Re: If the greatest common divisor of (n+2)!, (n-1)!, and (n+4)! is 120, w   [#permalink] 04 Feb 2019, 07:30
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