Latest self Aptitude Question SOLUTION: If n is a positive integer such that n!/(n-2)! = 342, find n.

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If n is a positive integer such that
n!/(n-2)! = 342, find n.

Devendra Marghade (12 years ago)

Solved Vignesh Read Solution (5)

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shilpi gupta  (12 years ago)

(n)(n-1)(n-2)!/(n-2)!=342
n^2-n-342=0
(n-19)(n+18)=0
n=19,-18

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Vignesh  (12 years ago)

n = 19

We write n! = n*(n-1)(n-2!)
Therefore
n!/(n-2)! = n(n-1)*(n-2)!/(n-2)! = n(n-1).

--> n(n-1) = 342
--> n^2 -n - 342 = 0
--> n^2- 19n +18n -342 = 0
--> n(n-19) +18(n -19) = 0
--> (n-19)(n+18) = 0

Therefore n -19 = 0; n +18 = 0;
(i.e) n =19 ; n = -18

We want positive integer. So, n=19.

19!/(19-2)! = 342
19!/17! = 342.

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Vignesh  (12 years ago)

n = 19

We write n! = n*(n-1)(n-2!)
Therefore
n!/(n-2)! = n(n-1)*(n-2)!/(n-2)! = n(n-1).

--> n(n-1) = 342
--> n^2 -n - 342 = 0
--> n^2- 19n +18n -342 = 0
--> n(n-19) +18(n -19) = 0
--> (n-19)(n+18) = 0

Therefore n -19 = 0; n +18 = 0;
(i.e) n =19 ; n =18

We want positive integer. So, n=19.

19!/(19-2)! = 342
19!/17! = 342.

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sonika thakur  (12 years ago)

ans: 19

n(n-1)(n-2)! / (n-2)!
n(n-1)=342
n^2 - n -342=0
n=-18, 19

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kirankumar M  (12 years ago)

(n*(n-1)*(n-2)!)/(n-2)!=342
n^2-n-342=0
n^2+19n-18n-342=0
n=19 (a positive integer)!

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