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
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 _________________
Sujoy Kumar Datta | GMAT - Q51 & CAT (MBA @ IIM) 99.98 Overall with 99.99 QA
IIT Kharagpur, TUD Germany
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