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Math Revolution and GMAT Club Contest! If the greatest common divisor

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Math Revolution and GMAT Club Contest! If the greatest common divisor [#permalink]

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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


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Re: Math Revolution and GMAT Club Contest! If the greatest common divisor [#permalink]

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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-
Since 5!=120 hence smallest of three numbers is 5.
Hence n-2=5 or n=7 Ans-D

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(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

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The greatest common divisor of (n+2)!, (n-2)! and (n+4)! is (n-2)!
Since 120 = 5! => n - 2 = 5 => n = 7
Answer D.
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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

By this question it simply means that 120 or 5! should be the largest common divisor among all the three nos.
Hence the value of n is 7 because 9!,5! and 11! has 5! as the largest common divisor.

D is the answer.
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New post 13 Dec 2015, 13:32
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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


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

The greatest common factor of above three is (n-2)! = 120 = 5!
So n-2 = 5 and n = 7

Answer (D)

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When considering the lowest common divisor of three factorials, we know that by the very nature of factorials, the (smallest number) factorial will be a divisor of any factorial of a larger number. This can be seen as \((x+1)!=x!*(x+1)\).

If 120 is the largest common divisor of these three factorials then we know that the value of the smallest factorial must equal 120. If this were not the case then the largest common divisor would be a different number.

The smallest factorial here is \((n-2)!\), and we know that \(5!=120\), therefore \(n-2=5\) and \(n=7\).

The answer is D.
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GCD of (n+2)!, (n-2)!, and (n+4)! is 120. Find n.

120 = 2^3 * 5 *3. Best way to find n is substitute each of the answer and find out the correct one. I started with the highest number.

Try 7 and find GCD or 9!, 5! and 11!. 5! itself is 120 and 5! which is contained in the other two factorials. This satisfies the condition, hence answer is 7. Also, we can eliminate all other answers as they are less than 7 and n-2 turns out to be less than 5! and hence cannot have a divisor of 120.

Answer is D.
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Looking at the factorials, n-2 is the smallest and 120 is the GCD

We need (n-2)! = 120
5! = 120
Hence n = 7

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GCF will be (n-2)!

thus (n-2)! = 120

as 5!=120

so, n=7

D

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Re: Math Revolution and GMAT Club Contest! If the greatest common divisor [#permalink]

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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"
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Re: Math Revolution and GMAT Club Contest! If the greatest common divisor [#permalink]

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120 = 2^3 * 3 *5
thus all 3 factorial numbers should include only 1 power of 3 and 5 and 3 powers of 2
==> n-2 will be the smallest number to include all these 3 means n-2 = 5 as if n-2 = 6 then 6! will include 3^2 and n+2 will also include 3^2 and so does n+4, therefore n-2 = 5 thus n = 7 to have GCD of 120 for (n+2)! , (n-2)! and (n+4)!.

Ans = D

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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

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Re: Math Revolution and GMAT Club Contest! If the greatest common divisor [#permalink]

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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

\(120 = 2^3*3*5=1*2*3*4*5=5!\)
In order for the 3 numbers to have the greatest common divisor 120 --> the smallest number (n-2)! at least has to equal 120.
--> (n-2)! = 5! --> n=7

Answer D

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Since (n-2)! is a factor of (n+2)! and (n+4)!, (n-2)! is the GCD of the 3 terms

We are given that numeric value of GCD is 5!

Therefore, (n-2)! = 5! => n = 7

Answer is D

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\(120 = 5!\). so for our 3 factorials to be divisible by 120, the smallest of the 3 must be at least equal to 5.
\(n-2 = 5 ==> n = 7\)
Answer D
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\(n\) = _____________ ?

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.

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option D.
Here the max value of GCD which is possible can be (n-2)! for all the numbers since this is the smallest among the three.

Now, both the numbers (n+2)! and (n+4)! can be written as (n-2)!*(something).
Thus GCD comes out to be (n-2)! which is given as 120.

Now since 120 = 5! => n-2 = 5.
Thus n=7.
option D is the answer.
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GCD for (n+2)!, (n-2)!, and (n+4)! would be (n-2)! which is equal to 120 as per the given data as 120 is 5!.
It can be written as (n-2)!=5!
=> n-2=5 and n=7

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Re: Math Revolution and GMAT Club Contest! If the greatest common divisor [#permalink]

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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
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