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find n

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find n

by jainrahul1985 » Mon Nov 07, 2011 5:06 am
If N is a positive integer and (n +1)! / (N-1)! = 72, find n ?
A. 6 B. 7 C. 8 D. 9 E. 10
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by Anurag@Gurome » Mon Nov 07, 2011 5:10 am
jainrahul1985 wrote:If N is a positive integer and (n +1)! / (N-1)! = 72, find n ?
A. 6 B. 7 C. 8 D. 9 E. 10
I hope N and n are the same variable, considering it as typo.

(n + 1)! /(n - 1)! = 72
(n + 1) * n * (n - 1)! /(n - 1)! = 72
n² + n - 72 = 0
n² + 9n - 8n - 72 = 0
n(n + 9) - 8(n + 9) = 0
n = 8

The correct answer is C.
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by GMATGuruNY » Mon Nov 07, 2011 5:17 am
jainrahul1985 wrote:If N is a positive integer and (n +1)! / (N-1)! = 72, find n ?
A. 6 B. 7 C. 8 D. 9 E. 10
We can plug in the answers, which represent the value of n.

Answer choice C: n=8.
(8+1)!/(8-1)!
= (9*8*7*6*5*4*3*2)/(7*6*5*4*3*2)
= 9*8
= 72.
Success!

The correct answer is C.
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by thestartupguy » Tue Nov 15, 2011 5:00 am
jainrahul1985 wrote:If N is a positive integer and (n +1)! / (N-1)! = 72, find n ?
A. 6 B. 7 C. 8 D. 9 E. 10
(n+1)!/(n-1)!= (n+1)(n)----(1)

72 = 9 X 8 ---(2)

comparing 1 and 2, we get n = 8
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by Abhishek009 » Tue Nov 15, 2011 7:29 am
jainrahul1985 wrote:If N is a positive integer and (n +1)! / (N-1)! = 72, find n ?
A. 6 B. 7 C. 8 D. 9 E. 10
Try using the options -


(n +1)! / (N-1)! = 72


We can arrive at the right hand side if we can , make( N + 1 )! as 9 * 8


For this we can take N as 8

So , ( 8 + 1 )! => 9! = 9*8*7*6....2*1

Again the denominator part....

We have to make keep the numerator as 9*8 and we must therefore remove the following part 7*6*5*4*3*2*1...

We can do it if we take ( N - 1)! as 7 ! and for that N must be 8....


So , N = 8
Abhishek
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