Fatorial #1

Algebra Level 2

If (x = n + 2), (y = 2n +2), (w = 2n + 1), (a = x - 1), (log 10 = 1)

and...

( x ) ! . ( y ) ! ( w ) ! . n ! . ( a ) \frac{(x)! . (y)!}{(w)! . n! . (a)} = log 1000 0 10 \log 10000^{10}

Find the possible value of n.

-6 6 -3 3

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1 solution

Victor Porto
Sep 9, 2014

[(x)! . (y)!] : [(w)! . n! . (a)] = log 10000^{10}

[(n + 2)! . (2n + 2)!] : [(2n + 1)! . n! (x - 1)] = 10 . log 10000

[(n + 2)! . (2n + 2)!] : [(2n + 1)! . n! (n + 2 - 1)] = 10 . log 10^{4}

[(n + 2)! . (2n + 2)!] : [(2n + 1)! . n! (n + 1)] = 4 . 10 . log 10

[(n + 2)! . (2n + 2)!] : [(2n + 1)! . n! (n + 1)] = 4 . 10 . 1

[(n + 2)! . (2n + 2)!] : [(2n + 1)! . n! (n + 1)] = 40

[(n + 2). (n +1) . (n!) . (2n + 2)!] : [(2n + 1)! . n! (n + 1)] = 40

[(n + 2) . (2n + 2)!] : [(2n + 1)!] = 40

[(n + 2) . (2n + 2) . (2n + 1)!] : [(2n + 1)!] = 40

[(n + 2) . (2n + 2)] = 40

(2n² + 2n + 4n + 4) = 40

2n² + 6n + 4 = 40

n² + 3n + 2 = 20

n² + 3n -18 = 0

n' = -6 n'' = 3

Note that the factorial of negative numbers is not defined, so n = 6 n = -6 is not a solution.

Only 3 is a solution.

I have updated the answer to 3.

Calvin Lin Staff - 6 years, 8 months ago

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Thanks @Calvin Lin !

Victor Porto - 6 years, 8 months ago

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