# Equation with factorials

by Alesak
Tags: equation, factorials
 P: 134 hello everyone, I usually post my questions on one small czech mathematical forum, but here is an equation noone knows how to "solve". Ive came to it by accident, when I made an mistake in one combinatorics equation. $$x! + (x-3)! = 16x - 24$$ its fairly simple to solve in one way(x has to be greater than 2, and from some point left side is greater than right side, because (x-3)! is always greater then 0 and we dont have to care about -24 on the right side, so we can check for which x is x! > 16x. this leaves us only very few possibilities for x to check). this is nice, but Id like to know if its possible to get it in form x = something. I cant think of any way how to do it. also, this equation doesn`t have any solution, its rather theoretical question.
 P: 11 If x can be a real number (using Gamma function) you can find 4 solutions betweeen 1 and 4.5 (by plotting method or std numeric method). x1=1.1837 x2=1.3134 x3=2.1222 x4=4.4099 There are also 5 negative solution... None is integer.
 P: 778 It is simple to check that there is no integer solution. Factorials increase far more rapidly than a linear function, more rapidly than exponentials even. We know x>2. For x=4, the LHS is smaller than the RHS, for x=5 it is larger, thus there is no integer x where they are equal.
P: 259

## Equation with factorials

I have also tried this in several different ways, I got:

x! = ( 8x (x-2) (x-1) (x-1.5) )/( x (x-1) ( x-2) +1 )

which means no integer values......

Is this really all????
 P: 11 Plotting the function (remebering x!=Gamma(x+1)) G(x)=Gamma(x + 1) + Gamma(x - 2) - 16x + 24 you can see that for x>0 there are only four zeros. For x<0 there are infinite solutions near 0,-1,-2,-3...none exacly integer. There Gamma function is singular (has a pole) and change sign.

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