National Mathematical Contest 2017!

Suppose $\log _5x=a$ , $\log _5y=b$ and $\log _5z=c$ , then $x\ =\ 5^a$ , $y\ =\ 5^b$ and $z\ =\ 5^c$

Hence, the system of equations becomes:

$5^{\left(a+b+c\right)}-3+a=32 \space [1]$

$5^{\left(a+b+c\right)}-3+b=81 \space [2]$

$5^{\left(a+b+c\right)}-3+c=256 \space [3]$

Adding the three equations:

$3\cdot 5^{\left(a+b+c\right)}-9+\left(a+b+c\right)=369$

Define $f\left(x\right)=3\cdot 5^x-9+x$ , then $f\left(a+b+c\right)=369=375-9+3=3\cdot 125-9+3=3\cdot 5^{\left(3\right)}-9+\left(3\right)=f\left(3\right)$

Note that, since $f(x)$ is monotonically increasing on $\mathbb{R}$ , it is a one-to-one function. Hence, since $f\left(a+b+c\right) = f(3)$ , $a+b+c = 3$ is the only possible solution. Furthermore, we subtract $[1]$ from $[2]$ and $[2]$ from $[3]$ and get the following:

$b - a = 49$

$c - b = 175$

And we also have $a + b + c = 3$ . Solving these three linear equations, we have $a = -90$ , $b = 134$ and $c = -41$ . Hence, we have $|\log _5x| + |\log _5y| + |\log _5z|= |a| + |b| + |c| = \boxed{265}$