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# If (n-2)! = (n! + (n-1)!)/99, and n is a positive integer, then n=?

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Joined: 13 Jul 2010
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If (n-2)! = (n! + (n-1)!)/99, and n is a positive integer, then n=? [#permalink]

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21 Nov 2010, 18:21
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If (n-2)! = (n! + (n-1)!)/99, and n is a positive integer, then n=?

A. 9
B. 10
C. 11
D. 99
E. 100
[Reveal] Spoiler: OA

Last edited by Bunuel on 16 Feb 2015, 02:50, edited 1 time in total.
Renamed the topic, edited the question and added the OA.
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Re: If (n-2)! = (n! + (n-1)!)/99, and n is a positive integer, then n=? [#permalink]

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21 Nov 2010, 20:33
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gettinit wrote:
If (n-2)!= $$n! + (n-1)! / 99$$, and n is a positive integer, then n=?

A. 9
b. 10
c. 11
d. 99
e. 100

I plugged-in to arrive at answer - anyone have a faster way? Also when plugging-in is there a recommended method for instance start with c (move up or down depending on answer) or start with E as GMAT knows your plugging in? thanks

I am assuming here that the question is (n-2)!= $$\frac{(n! + (n-1)!)}{99}$$. I wish you would put brackets for clarity (you anyway take the pains of putting the m tag)
$$(n-2)!= \frac{(n - 2)!(n(n - 1) + (n-1))}{99}$$ (Taking (n - 2)! common out of n! and (n - 1)!)
$$99 = n^2 - 1$$
n = 10

Remember, when dealing with multiple factorials, all you can do is take something common.

When you try options, try the value in the middle (e.g. if you have 10, 20, 30, 40 and 50, try 30 first. It will tell you whether you need to go up or down i.e. whether you need to try 10/20 or 40/50. You will definitely save some steps)
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Manager Status: Still Struggling Joined: 02 Nov 2010 Posts: 132 Location: India GMAT Date: 10-15-2011 GPA: 3.71 WE: Information Technology (Computer Software) Re: If (n-2)! = (n! + (n-1)!)/99, and n is a positive integer, then n=? [#permalink] ### Show Tags 21 Nov 2010, 22:22 VeritasPrepKarishma wrote: gettinit wrote: If (n-2)!= $$n! + (n-1)! / 99$$, and n is a positive integer, then n=? A. 9 b. 10 c. 11 d. 99 e. 100 I plugged-in to arrive at answer - anyone have a faster way? Also when plugging-in is there a recommended method for instance start with c (move up or down depending on answer) or start with E as GMAT knows your plugging in? thanks I am assuming here that the question is (n-2)!= $$\frac{(n! + (n-1)!)}{99}$$. I wish you would put brackets for clarity (you anyway take the pains of putting the m tag) $$(n-2)!= \frac{(n - 2)!(n(n - 1) + (n-1))}{99}$$ (Taking (n - 2)! common out of n! and (n - 1)!)$$99 = n^2 - 1$$ n = 10 Remember, when dealing with multiple factorials, all you can do is take something common. When you try options, try the value in the middle (e.g. if you have 10, 20, 30, 40 and 50, try 30 first. It will tell you whether you need to go up or down i.e. whether you need to try 10/20 or 40/50. You will definitely save some steps) I didnt understand how you took (n-2)! out of the RHS. Can you please explain in detail? Thanks, ----------------------------------------------------------------------- Consider KUDOS if you like my post!! _________________ Appreciation in KUDOS please! Knewton Free Test 10/03 - 710 (49/37) Princeton Free Test 10/08 - 610 (44/31) Kaplan Test 1- 10/10 - 630 Veritas Prep- 10/11 - 630 (42/37) MGMAT 1 - 10/12 - 680 (45/34) Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7944 Location: Pune, India Re: If (n-2)! = (n! + (n-1)!)/99, and n is a positive integer, then n=? [#permalink] ### Show Tags 22 Nov 2010, 04:45 3 This post received KUDOS Expert's post 3 This post was BOOKMARKED Sure. Lets take an example first: 2! = 1*2 3! = 1*2*3 = 2!*3 (Since 1*2 = 2!, substitute it here) 4! = 1*2*3*4 = 2!*3*4 or 3!*4 So, n! = (n - 1)! * n Also, n! = (n - 2)! * (n - 1) * n n! = (n - 3)! * (n - 2) * (n - 1) * n and so on... Therefore, when we have (n-1)! + n!, we can write this as (n - 1)! = (n - 2)! * (n -1) n! = (n - 2)! * (n - 1) * n So (n-1)! + n! = (n - 2)! * (n -1) + (n - 2)! * (n - 1) * n (n-1)! + n! = (n - 2)! [(n - 1) + (n -1)*n] Note: We could have taken (n - 1)! common but we only need (n - 2)! because only (n - 2)! gets canceled in the question above. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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Re: If (n-2)! = (n! + (n-1)!)/99, and n is a positive integer, then n=? [#permalink]

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22 Nov 2010, 20:18
Karishma, sorry about the parentheses I'll be sure to add in the future still trying to get a hold of this thing. Thanks for the reminder.

Question on your factoring of n! and (n-1)!, how do you know that n!=(n-2)!*(n-1)!*n! why could this not continue from (n-3)!..(n-4)! etc. My question is we know 2! is 2*1 but we don't know what n! factorial is so how can we determine n! factorial is factored a certain way while (n-1)! another? Thanks for the clarification
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Re: If (n-2)! = (n! + (n-1)!)/99, and n is a positive integer, then n=? [#permalink]

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22 Nov 2010, 21:32
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gettinit wrote:
If (n-2)!= $$n! + (n-1)! / 99$$, and n is a positive integer, then n=?

A. 9
b. 10
c. 11
d. 99
e. 100

I plugged-in to arrive at answer - anyone have a faster way? Also when plugging-in is there a recommended method for instance start with c (move up or down depending on answer) or start with E as GMAT knows your plugging in? thanks

Note that n! = n * (n-1)! = n * (n-1) * (n-2)! (By definition of factorials)

(n-2)! = n(n-1)(n-2)!/99 + (n-1)(n-2)!/99
1 = n(n-1)/99 + (n-1)/99
99 = n^2 -n + n - 1
n^2 - 100 = 0

Hence, n=10
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Re: If (n-2)! = (n! + (n-1)!)/99, and n is a positive integer, then n=? [#permalink]

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23 Nov 2010, 04:57
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gettinit wrote:
Karishma, sorry about the parentheses I'll be sure to add in the future still trying to get a hold of this thing. Thanks for the reminder.

Question on your factoring of n! and (n-1)!, how do you know that n!=(n-2)!*(n-1)!*n! why could this not continue from (n-3)!..(n-4)! etc. My question is we know 2! is 2*1 but we don't know what n! factorial is so how can we determine n! factorial is factored a certain way while (n-1)! another? Thanks for the clarification

n! = 1*2*3*4*5*6*...*(n - 2)*(n -1)*n

From above, can I say that n! = (1*2*3*4)*5*6*...*n = 4!*5*6*...*n
(Assuming that n is greater than 8)

Can I also say, n! = (1*2*3*4*5*6...*(n - 2))*(n - 1)*n = (n - 2)!*(n -1)*n

e.g. if n = 10
10! = 1*2*3*4*5*6*7*8*9*10 = 2!*3*4*5*6*7*8*9*10 = 3!*4*5*6*7*8*9*10 = 4!*5*6*7*8*9*10 = 5!*6*7*8*9*10 = 6!*7*8*9*10 = 7!*8*9*10 = 8!*9*10 = 9!*10
So n! can be (n-1)!*n or (n-2)!*(n - 1)*n and so on...

We can write n! in many different ways. We use whatever suits us best in the question. Here, since the left hand side has (n - 2)!, we club the first (n - 2) numbers together and write them as (n - 2)! and let the last two (n -1) and n be. Since we can take (n - 2)! from both n! and (n - 1)!, we can cancel it out and are left with a simple quadratic.
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Manager Joined: 13 Jul 2010 Posts: 159 Re: If (n-2)! = (n! + (n-1)!)/99, and n is a positive integer, then n=? [#permalink] ### Show Tags 23 Nov 2010, 18:25 Karishma, I get it thanks for the excellent explanation. appreciate your time. Kudos! Intern Joined: 25 Nov 2009 Posts: 42 Location: India Re: If (n-2)! = (n! + (n-1)!)/99, and n is a positive integer, then n=? [#permalink] ### Show Tags 23 Nov 2010, 21:33 I took a hint from 99 (9*11) and b'coz of addition tried n=10 first. _________________ When going gets tough, tough gets going......... Manager Joined: 19 Aug 2010 Posts: 71 Re: If (n-2)! = (n! + (n-1)!)/99, and n is a positive integer, then n=? [#permalink] ### Show Tags 07 Feb 2011, 13:36 VeritasPrepKarishma wrote: gettinit wrote: If (n-2)!= $$n! + (n-1)! / 99$$, and n is a positive integer, then n=? I am assuming here that the question is (n-2)!= $$\frac{(n! + (n-1)!)}{99}$$. I wish you would put brackets for clarity (you anyway take the pains of putting the m tag) $$(n-2)!= \frac{(n - 2)!(n(n - 1) + (n-1))}{99}$$ (Taking (n - 2)! common out of n! and (n - 1)!) $$99 = n^2 - 1$$ n = 10 ) Оххх, thanks many times. I lost so much time ... Non-Human User Joined: 09 Sep 2013 Posts: 13810 Re: If (n-2)! = (n! + (n-1)!)/99, and n is a positive integer, then n=? [#permalink] ### Show Tags 14 Feb 2015, 00:33 Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________ EMPOWERgmat Instructor Status: GMAT Assassin/Co-Founder Affiliations: EMPOWERgmat Joined: 19 Dec 2014 Posts: 11046 Location: United States (CA) GMAT 1: 800 Q51 V49 GRE 1: 340 Q170 V170 Re: If (n-2)! = (n! + (n-1)!)/99, and n is a positive integer, then n=? [#permalink] ### Show Tags 14 Feb 2015, 13:02 1 This post received KUDOS Expert's post Hi Mods, Could one of you put brackets into the original prompt (as described in the follow-up posts) - it would save any future Users the trouble of working on an incorrectly-formatted question. Thanks, GMAT assassins aren't born, they're made, Rich _________________ 760+: Learn What GMAT Assassins Do to Score at the Highest Levels Contact Rich at: Rich.C@empowergmat.com # Rich Cohen Co-Founder & GMAT Assassin Special Offer: Save$75 + GMAT Club Tests Free
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Re: If (n-2)! = (n! + (n-1)!)/99, and n is a positive integer, then n=? [#permalink]

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16 Feb 2015, 02:50
EMPOWERgmatRichC wrote:
Hi Mods,

Could one of you put brackets into the original prompt (as described in the follow-up posts) - it would save any future Users the trouble of working on an incorrectly-formatted question.

Thanks,

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Rich

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Done. Thank you.
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Re: If (n-2)! = (n! + (n-1)!)/99, and n is a positive integer, then n=? [#permalink]

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16 Feb 2015, 06:20
Plugged in values.....took 1 min....Luckily option B it is
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Re: If (n-2)! = (n! + (n-1)!)/99, and n is a positive integer, then n=? [#permalink]

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28 Mar 2016, 17:49
Hello from the GMAT Club BumpBot!

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Re: If (n-2)! = (n! + (n-1)!)/99, and n is a positive integer, then n=? [#permalink]

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18 Apr 2017, 01:47
Option B

(n-2)! = (n! + (n-1)!)/99

(n-2)! = (n+1)(n-1)!/99

99 = (n+1)(n-1)

n = 10.
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Re: If (n-2)! = (n! + (n-1)!)/99, and n is a positive integer, then n=? [#permalink]

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23 Apr 2017, 18:52
shrouded1 wrote:
gettinit wrote:
If (n-2)!= $$n! + (n-1)! / 99$$, and n is a positive integer, then n=?

A. 9
b. 10
c. 11
d. 99
e. 100

I plugged-in to arrive at answer - anyone have a faster way? Also when plugging-in is there a recommended method for instance start with c (move up or down depending on answer) or start with E as GMAT knows your plugging in? thanks

Note that n! = n * (n-1)! = n * (n-1) * (n-2)! (By definition of factorials)

(n-2)! = n(n-1)(n-2)!/99 + (n-1)(n-2)!/99
1 = n(n-1)/99 + (n-1)/99
99 = n^2 -n + n - 1
n^2 - 100 = 0

Hence, n=10

Thank You- this is a clear explanation - is it necessary to set it equal to 0 though? Couldn't we just leave it at n^2-1=99 - or maybe I suppose a faster/better mathematician just does it so efficiently that they arrive at that step.
Re: If (n-2)! = (n! + (n-1)!)/99, and n is a positive integer, then n=?   [#permalink] 23 Apr 2017, 18:52
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