$$100!=K(2!)^x(5!)^y(7!)^z$$ Find the maximum value of $x+y+z$.
I tried to open all the brackets and I got $$100!=K(2)^{x+3y+4z}(3)^{y+2z}(5)^{y+z}(7)^{z}$$
By using greatest integer function, I got the following inequalities: $$x+3y+4z\le97$$ $$y+2z\le48$$ $$y+z\le24$$ $$z\le16$$ By adding the last two, I got $$y+2z\le40$$ Now I can't figure out how to proceed from here.
Edit: $x, y$ and $z$ are non-negative integers.
If $x,y,z$ are nonnegative integers, as I suppose, then the answer is $97$, achieved when $x=97, y=z=0$.
Just look at the first inequality. If we decrease $z$ by $1$, we can increase $x$ by $4$, giving a net increase of $3$ to $x+y+z$. Similarly, if we decrease $y$ by $1$, we can increase $x+y+z$ by $2$.