algebra (nice factorial question I made up) Watch

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john !!
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#1
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#1
2x[(x-1)!]^2 = 6x! - 36x

x is a natural number.
n! is the product of all the natural numbers from 1 to n inclusive.

Find the value of x.

this is probably a good challenge to A level people.
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Gaz031
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(Original post by mik1w)
2x[(x-1)!]^2 = 6x! - 36x

x is a natural number.
n! is the product of all the natural numbers from 1 to n inclusive.

Find the value of x.

this is probably a good challenge to A level people.
2x[(x-1)!]^2 = 6x! - 36x
2x[ (x-1)! ]^2 = 6x(x-1)! - 36x
2[ (x-1)! ]^2 = 6(x-1)! - 36
If y=(x-1)!:
2y^2 = 6y - 36
2y^2 - 6y + 36 = 0
y^2 - 3y + 18 = 0
No real roots.
Erm? Have you ensured there is an answer?
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john !!
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2x[(x-1)!]^2 = 36x + 6x!

sorry I think I meant this^!
always missing signs :-\
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Gaz031
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2x[(x-1)!]^2 = 36x + 6x!
2[(x-1)!]^2 = 36 + 6(x-1)!
y = (x-1)!
2y^2 - 6y - 36 = 0
y^2 - 3y - 18 = 0
y = (3+-9)/2
We take the positive, hence y=6

(x-1)! = 6
x=4.


That seems to work out better. Interesting question.
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C4>O7
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2x[(x-1)!]^2 = 36x + 6x!
2[(x-1)!]^2 = 36 + 6(x-1)!
Let a = (x-1)!
a^2 - 3a - 18 = 0
(a-6)(a+3) = 0
(x-1)! = 6 or -3
Reject (x-1)! = -3 as x is natural
-> (x-1)! = 6
-> x = 4
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