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If n!/(n-2)!<100, what is the greatest possible value of

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by Max@Math Revolution » Mon Mar 05, 2018 12:37 am
[GMAT math practice question]

If n!/(n-2)!<100, what is the greatest possible value of n?

A. 8
B. 9
C. 10
D. 11
E. 12
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by mbawisdom » Mon Mar 05, 2018 7:14 am
Max@Math Revolution wrote:[GMAT math practice question]

If n!/(n-2)!<100, what is the greatest possible value of n?

A. 8
B. 9
C. 10
D. 11
E. 12
n!/(n-2)! < 100
(n*(n-1)*(n-2)!)/(n-2)! < 100
n*(n-1) <100

E) n of 12 doesn't hold as 121 >= 100
D) n of 11 doesn't hold as 110 >= 100
C) n of 10 does hold as 90 < 100
No need to test B and A

Answer is C. 10
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by Max@Math Revolution » Wed Mar 07, 2018 2:31 am
=>

We must have
n! / (n-2)! = n(n-1) < 100.
If n = 10, n(n-1) = 10*9 = 90 < 100.
If n = 11, n(n-1) = 11*10 = 110 > 100.

10 is the greatest value of n for which n!/(n-2)! < 100.

Therefore, C is the answer.

Answer: C
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