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

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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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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I didnt understand how you took (n-2)! out of the RHS. Can you please explain in detail?

Thanks,

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Consider KUDOS if you like my post!!

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

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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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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Karishma, I get it thanks for the excellent explanation. appreciate your time. Kudos!

I took a hint from 99 (9*11) and b'coz of addition tried n=10 first.

Оххх, thanks many times. I lost so much time ...

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

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Done. Thank you.
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Plugged in values.....took 1 min....Luckily option B it is

Option B

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

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

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

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.

We have:

\((n-2)! = \frac{(n! + (n-1)!)}{99}\)

\(=> 99 * (n-2)! = n! + (n-1)!\)


Considering factorials, we have:

\(n! = 1 * 2 * 3 * ... * (n - 2) * (n - 1) * n\)

\(=> n! = [1 * 2 * 3 * ... * (n - 2)] * (n - 1) * n\)

\(=> n! = (n - 2)! * (n - 1) * n\)

Similarly: \(=> (n - 1)! = (n - 2)! * (n - 1)\)

\(=> n! + (n - 1)! = [(n - 2)! * (n - 1)] * n + [(n - 2)! * (n - 1)]\)

\(= [(n - 2)! * (n - 1)] * (n + 1)\)


Thus, we have: \(99 * (n-2)! = n! + (n-1)!\)

\(=> 99 * (n-2)! = [(n - 2)! * (n - 1)] * (n + 1)\)

Cancelling \((n - 2)!\) from both sides:

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

\(=> 99 = n^2 - 1\)

\(=> n = 10\) (considering the positive value of n)

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

\(n! + (n-1)! = (n-1)! (n + 1) = (n-2)! (n-1)(n+1) = (n-2)! (nˆ2 - 1) = 99 (n-2)!\)
\(nˆ2 -1=99\)
\(nˆ2 = 100\)
n = 10

IMO B