Question 1206590
.
If 3n³× (n-2)! + 6n - 3n² × (n-3)! - 360 = 0 , find value of n ?
~~~~~~~~~~~~~~~~


<pre>
  {{{3n^3(n-2)!}}} + {{{6n}}} - {{{3n^2*(n-3)!}}} - 360 =

= {{{3n^2}}}.{{{(n*(n-2)-1)}}}.{{{(n-3)!}}} + 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*4^2}}}.{{{(4*(4-2)-1)}}}.{{{(4-3)!}}} + 6*4 - 360 = 3*16*7*1 + 24 - 360 = 0.


So, the <U>ANSWER</U> to the problem is n= 4.
</pre>

Solved.


-------------------


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.