SOLUTION: If 3n³× (n-2)! + 6n - 3n² × (n-3)! - 360 = 0 , find value of n ?

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Question 1206590: If 3n³× (n-2)! + 6n - 3n² × (n-3)! - 360 = 0 , find value of n ?
Answer by ikleyn(52787) About Me  (Show Source):
You can put this solution on YOUR website!
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If 3n³× (n-2)! + 6n - 3n² × (n-3)! - 360 = 0 , find value of n ?
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  3n%5E3%28n-2%29%21 + 6n - 3n%5E2%2A%28n-3%29%21 - 360 =

= 3n%5E2.%28n%2A%28n-2%29-1%29.%28n-3%29%21 + 6n - 360.


Now you can see that this expression is monotonically growing function of n.

So, if we guess one solution, it will provide a UNIQUE solution.


It is not difficult to guess one solution. It is n= 4, because

    3%2A4%5E2.%284%2A%284-2%29-1%29.%284-3%29%21 + 6*4 - 360 = 3*16*7*1 + 24 - 360 = 0.


So, the ANSWER to the problem is n= 4.

Solved.

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In such problems, guessing of a root in combinations with monotonicity
of a function turns guessing in strict mathematical proof.

In other words, in this combination, guessing of a root acquires
the force and the status of a valid mathematical proof.