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If the greatest common divisor

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by GMATGuruNY » Fri Jan 12, 2018 3:42 am
lheiannie07 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
We can PLUG IN THE ANSWERS, which represent the value of n.
120 = 2*3*4*5 = 5!.
Since 5! is the GCF of the three given values, (n-1)! must be at least 5!.
Eliminate A, B and E, which will render too small a value for (n-1)!.

D: n=7
Here, (n+2)! = 9!, (n-1)! = 6!, and (n+4)! = 11!, with the result that all three values are divisible by 6!.
Since the GCF must be less than 6!, eliminate D.

The correct answer is C.
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by [email protected] » Fri Jan 12, 2018 11:37 am
Hi lheiannie07,

We're told that the greatest common divisor of (N+2)!, (N-1)!, and (N+4)! is 120. We're asked for the value of N. This question can be approached in a number of different ways. Here's how you can use a bit of logic and Prime Factorization to get the answer.

To start, it's worth noting that a larger factorial will be divisible by ALL of the factors that appear in a smaller factorial. For example: 4! = 24 and 3! = 6. ALL of the factors of 3! (re: 1, 2, 3 and 6) are also factors of 4! (re: 1, 2, 3, 4, 6, 8, 12, and 24). Thus, (N+2)! and (N+4)! are both bigger than (N-1)! so they will both be divisible by the same factors as (N-1)! This is meant to say that to get the answer, we just have to make sure that we make the N as small as possible AND that (N-1)! is divisible by 120.

If we 'prime factor' 120, we get....
120 =
(10)(12) =
(2)(5)(2)(2)(3)

For (N-1)! to be divisible by 120, it has to be divisible by 5... which means that '5' has to be a part of the factorial calculation. The smallest value for N in which that will happen is when N=6:
(6-1)! = 5! = (5)(4)(3)(2)(1)

Final Answer: C

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by BTGmoderatorDC » Sun Jan 14, 2018 4:38 pm
GMATGuruNY wrote:
lheiannie07 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
We can PLUG IN THE ANSWERS, which represent the value of n.
120 = 2*3*4*5 = 5!.
Since 5! is the GCF of the three given values, (n-1)! must be at least 5!.
Eliminate A, B and E, which will render too small a value for (n-1)!.

D: n=7
Here, (n+2)! = 9!, (n-1)! = 6!, and (n+4)! = 11!, with the result that all three values are divisible by 6!.
Since the GCF must be less than 6!, eliminate D.

The correct answer is C.
Thanks a lot!
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by BTGmoderatorDC » Sun Jan 14, 2018 4:43 pm
[email protected] wrote:Hi lheiannie07,

We're told that the greatest common divisor of (N+2)!, (N-1)!, and (N+4)! is 120. We're asked for the value of N. This question can be approached in a number of different ways. Here's how you can use a bit of logic and Prime Factorization to get the answer.

To start, it's worth noting that a larger factorial will be divisible by ALL of the factors that appear in a smaller factorial. For example: 4! = 24 and 3! = 6. ALL of the factors of 3! (re: 1, 2, 3 and 6) are also factors of 4! (re: 1, 2, 3, 4, 6, 8, 12, and 24). Thus, (N+2)! and (N+4)! are both bigger than (N-1)! so they will both be divisible by the same factors as (N-1)! This is meant to say that to get the answer, we just have to make sure that we make the N as small as possible AND that (N-1)! is divisible by 120.

If we 'prime factor' 120, we get....
120 =
(10)(12) =
(2)(5)(2)(2)(3)

For (N-1)! to be divisible by 120, it has to be divisible by 5... which means that '5' has to be a part of the factorial calculation. The smallest value for N in which that will happen is when N=6:
(6-1)! = 5! = (5)(4)(3)(2)(1)

Final Answer: C

GMAT assassins aren't born, they're made,
Rich
Thanks a lot!
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TTT

by DrMaths » Mon Jan 15, 2018 4:06 am
Note that 120 = 5 x 4 x 3 x 2 x 1 = 5!
Also, (n-1)! is the smallest value that requires the greatest divisor, so we don't want anything bigger than that.
So, if we let (n-1)! = 5!, we get n = 6.
Answer C
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by Scott@TargetTestPrep » Mon Aug 05, 2019 6:16 pm
BTGmoderatorDC 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

Can some experts show me solution in this problem?

OA C

We need the following fact:

If n < m, then n! is a divisor of m!.

For example, 4! is a divisor of 5! and 3! is a divisor of 8!. Now, since 5! = 120, we need n to be at least 6, in order to make (n - 1)! divisible by 5!. So this eliminates choices A, B, and E. Let's check the remaining two choices:

If n = 6, then (n+2)! = 8!, (n-1)! = 5!, and (n+4)! = 9!. Since 5! Is the smallest of them, 5! = 120 would be the greatest common divisor.

If n = 7, then (n+2)! = 9!, (n-1)! = 6!, and (n+4)! = 10!. Since 6! Is the smallest of them, 6! = 720 would be the greatest common divisor.

Therefore, we see that n must be 6.

Alternate Solution:

Since (n - 1)! is a divisor of both (n + 2)! and (n + 4)!; GCD of (n + 4)!, (n + 2)! and (n - 1)! is (n - 1)!. We are told that the GCD of these three expressions is 120, thus (n - 1)! = 120. Since 120 = 5!, we have n - 1 = 5 and so, n = 6.

Answer: C

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