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

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

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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

Originally posted by gettinit on 21 Nov 2010, 19:21.
Last edited by Bunuel on 16 Feb 2015, 03:50, 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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New post 21 Nov 2010, 21:33
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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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Re: If (n-2)! = (n! + (n-1)!)/99, and n is a positive integer, then n=? [#permalink]

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New post 21 Nov 2010, 23:22
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,

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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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New post 22 Nov 2010, 05:45
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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.
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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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New post 22 Nov 2010, 21:18
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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New post 22 Nov 2010, 22:32
2
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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New post 23 Nov 2010, 05:57
2
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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Re: If (n-2)! = (n! + (n-1)!)/99, and n is a positive integer, then n=? [#permalink]

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New post 23 Nov 2010, 19:25
Karishma, I get it thanks for the excellent explanation. appreciate your time. Kudos!
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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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New post 23 Nov 2010, 22:33
I took a hint from 99 (9*11) and b'coz of addition tried n=10 first.
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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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New post 07 Feb 2011, 14:36
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 ...
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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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New post 14 Feb 2015, 14:02
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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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Re: If (n-2)! = (n! + (n-1)!)/99, and n is a positive integer, then n=? [#permalink]

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New post 16 Feb 2015, 03:50
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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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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New post 16 Feb 2015, 07:20
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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New post 18 Apr 2017, 02:47
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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New post 23 Apr 2017, 19:52
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, 19:52
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