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

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

GCD=common factors=(n-2)!
Given, GCD=120
n-2=5=>n=7
Ans D

If the greatest common divisor of (n+2)!, (n-2)!, and (n+4)! is 120, what is the value of n?

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

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

hereby, Greatest common divisor= (n-2)!

Given condition: (n-2)! will be 120

if we plug (n-2)! into answers, we get:

A. 4 =(4-2)!=2!=2
B. 5 =(5-2)!=3!=6
C. 6 =(6-2)!=4!=24
D. 7 =(7-2)!=5!=120
E. 3=(3-2)!=1!=1


Answer: "D"

greatest common divisor of (n+2)!, (n-2)!, and (n+4)! is 120

Each of (n+2)!, (n-2)!, and (n+4)! has to be divisible by 120

120 = 5!

(n+2)! has to be divisible by 5! --> n can be an integer >= 3
(n-2)! has to be divisible by 5! --> n can be an integer >= 7
(n+4)! has to be divisible by 5! --> n can be an integer >= 1

From the above values of n it can be seen that the minimum value of n has to be 7.

Answer: D

\(n\) = _____________ ?

Arrange this expression in order: \(( n - 2 ) !\) , \(( n + 2 ) !\) and \(( n + 4 ) !\).

Since these are factorials, it means that \(( n - 2 ) !\) is contained within both \(( n + 2 )!\) and \(( n + 4 ) !\). So the greatest common factor (or divisor) of all these expressions is actually \(( n - 2 ) !\).

\(( n - 2 ) ! = 120\) --> \(( n - 2 ) ! = 5 !\) --> \(n = 7\).


The demonstration :

\(( n - 2 ) ! = ( n - 2 ) * ( n - 3 ) * ( n - 4 ) * ( n - 5 ) * (....)\). We don't know where this factorization stops, it depends on the value of n. But we don't need to know.

\(( n + 2 ) ! = ( n + 2 ) * ( n + 1 ) * ( n ) * ( n - 1 ) * ( n - 2 ) !\)

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

So as you can see \(( n - 2 ) !\) is actually the greatest common factor for all three expressions. You can be sure that it is the greatest common factor because none of the factors of ( n + 2 ) ! and ( n + 4) ! can be found in the factors of ( n - 2) ! .

I hope that this is clear.

Answer D.

QUESTION #11:

If the greatest common divisor of (n+2)!, (n-2)!, and (n+4)! is 120, what is the value of n?

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

--
Expanding :
(n+2)! , (n-2)! and (n+4)! :

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

Each of these factorials have a common factorial (n-2)! , so (n-2)! is one possible divisor.
Given that a largest factor for any integer K is K itself, AND that we're looking for the GREATEST COMMON factor for the three factorials presented, it can be deduced that (n-2)! is indeed the greatest common factor looking for.
any factor bigger than (n-2)! cannot be common across the three terms because it cannot divide (n-2)! - so cannot be common ,
and any factor of the three terms , lesser than (n-2)! cannot be the GCD because it cannot be greatest.

Hence, (n-2)! is the GCD.

Given : (n-2)! = 120 => n-2 = 5
Hence n = 7;

Answer : D

MATH REVOLUTION OFFICIAL SOLUTION:

\((n+2)!=(n+2)(n+1)n(n-1)(n-2)!\) and \((n+4)!=(n+4)(n+3)(n+2)(n+1)n(n-1)(n-2)!\).

From this, we can observe that (n+4)!, (n+2)! and (n-2)! include (n-2)!, which means the greatest common divisor for them is (n-2)!. So, since we get n-2=5 and n=7 from \((n-2)!=120=5!\), the correct answer is D.

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