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

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Math Revolution GMAT Instructor
Joined: 16 Aug 2015
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If n!/(n-2)!<100, what is the greatest possible value of n? [#permalink]

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05 Mar 2018, 01:41
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Difficulty:

5% (low)

Question Stats:

82% (00:30) correct 18% (00:39) wrong based on 65 sessions

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[GMAT math practice question]

If $$\frac{n!}{(n-2)!}<100$$, what is the greatest possible value of $$n$$?

$$A. 8$$
$$B. 9$$
$$C. 10$$
$$D. 11$$
$$E. 12$$

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"Only $99 for 3 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" Manager Joined: 31 Oct 2013 Posts: 236 Concentration: Accounting, Finance GPA: 3.68 WE: Analyst (Accounting) Re: If n!/(n-2)!<100, what is the greatest possible value of n? [#permalink] ### Show Tags 05 Mar 2018, 03:14 MathRevolution wrote: [GMAT math practice question] If $$\frac{n!}{(n-2)!}<100$$, what is the greatest possible value of $$n$$? $$A. 8$$ $$B. 9$$ $$C. 10$$ $$D. 11$$ $$E. 12$$ i would like to use back solving strategy. ! but before doing so , a simple thing that must be mentioned is that numerator is n! and denominator is (n-2)! it means that 2 no. will always be extra in numerator part. how? a) 8 as a value of n so, 8!/ (8-2)! = 8!/6! = 8*7*6*5*4*3*2*1/6*5*4*3*2*1 = 8*7 =56 <100 , it meets our condition . now , we have to remember that we are looking for max value of n it must be 10 10!/(10-2)! =10*9*8*7*6*5*4*3*2*1/8*7*6*5*4*3*2*1 =10*9 =90 <100 the correct answer is C. Manager Joined: 02 Jan 2016 Posts: 82 Re: If n!/(n-2)!<100, what is the greatest possible value of n? [#permalink] ### Show Tags 05 Mar 2018, 20:53 MathRevolution wrote: [GMAT math practice question] If $$\frac{n!}{(n-2)!}<100$$, what is the greatest possible value of $$n$$? $$A. 8$$ $$B. 9$$ $$C. 10$$ $$D. 11$$ $$E. 12$$ n!/(n-2)!, will be the last two factors of N!, and it has to be such that it is less than 100, 9*10 = 90, less than 100, if we go further 10*11 = 110 this is beyond the restriction. thus 10! is the answer Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 5600 GMAT 1: 800 Q59 V59 GPA: 3.82 Re: If n!/(n-2)!<100, what is the greatest possible value of n? [#permalink] ### Show Tags 07 Mar 2018, 03:39 => We must have $$\frac{n!}{(n-2)!} = n(n-1) < 100.$$ 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 $$\frac{n!}{(n-2)!} < 100.$$ Therefore, C is the answer. Answer: C _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only$99 for 3 month Online Course"
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Manager
Joined: 23 May 2017
Posts: 226
Concentration: Finance, Accounting
WE: Programming (Energy and Utilities)
Re: If n!/(n-2)!<100, what is the greatest possible value of n? [#permalink]

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07 Mar 2018, 04:15
Equation can be reduced to n ( n - 1) < 100

Best way is to "Test the Numbers"

Lets pick from the largest number -

[E] : 12 * 11 = 132 < 100 : No
[D] : 11 * 10 = 110 < 100 : No
[C] : 10 * 9 = 90 < 100 : Yes - This is the answer
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Math Revolution GMAT Instructor
Joined: 16 Aug 2015
Posts: 5600
GMAT 1: 800 Q59 V59
GPA: 3.82
Re: If n!/(n-2)!<100, what is the greatest possible value of n? [#permalink]

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21 Mar 2018, 11:29
Leo8 wrote:
Equation can be reduced to n ( n - 1) < 100

Best way is to "Test the Numbers"

Lets pick from the largest number -

[E] : 12 * 11 = 132 < 100 : No
[D] : 11 * 10 = 110 < 100 : No
[C] : 10 * 9 = 90 < 100 : Yes - This is the answer

For this question, your way is the best one.
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Re: If n!/(n-2)!<100, what is the greatest possible value of n?   [#permalink] 21 Mar 2018, 11:29
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