Problem Solving (PS) | If (n + 2)!/n! = 90, then what is the value of positive integer n ?

Breaking down an algebraic factorial:

If we understand that (4)! = (4) x (3) x (2) x (1) ------> this is the same as (4 - 0) x (4 - 1) x (4 - 2) x (4 - 3)

then by extension, (n)! = (n - 0) x (n - 1) x (n - 2) .... etc

and, (n + 2)! = (n + 2) x (n + 1) x (n - 0) x (n - 1) x (n - 2) ... etc

therefore:

(n + 2)! / (n)! = (n + 2) x (n + 1)

= n^2 + 3n + 2

and since n^2 + 3n + 2 = 90, we can set this equation to 0 by moving 90 to the quadratic equation side

n^2 + 3n - 88 = 0. Now we can solve the quadratic equation and get

(n + 11) (n - 8)

Result:
n = - 11
n = 8

Since the question is looking for "positive integer value of n"

n = 8 is the only answer.

Statistics : Posted by GKLemon • on 31 Jan 2023, 07:24 • Replies 1 • Views 192

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