Summing powers

Without loss of generality , let $x> y> z$ .

$\large{3^x+3^y+3^z=591219 \\ \Rightarrow 3^z(3^{x-z}+3^{y-z}+1)=3^6 \times 811}$

Since $\large{(3^{x-z}+3^{y-z}+1)}$ is not divisible by $3$ , no power of $3$ must be assigned to it. Hence by comparing $L.H.S \ , \ R.H.S$ , we get :

$\large{3^z=3^6 \Rightarrow \boxed{z=6} \ ; \ 3^{x-z}+3^{y-z}+1= 811 \\ \Rightarrow 3^{x-6}+3^{y-6}= 810 \\ \Rightarrow 3^{y-6}(3^{x-y}+1)=3^4 \times 10}$

Since $\large{(3^{x-y}+1)}$ is not divisible by $3$ , no power of $3$ must be assigned to it. Hence by comparing $L.H.S \ , \ R.H.S$ , we get :

$\large{3^{y-6}=3^4 \Rightarrow y-6=4 \Rightarrow \boxed{y=10} \\ Also \ , \ (3^{x-y}+1) = 10 \\ \Rightarrow 3^{x-10} = 9 \Rightarrow 3^{x-10}=3^2 \Rightarrow x-10 = 2 \Rightarrow \boxed{x=12}}$

Thus , finally , $\Large{x+y+z=6+10+12=\boxed{28}}$ .

This is similar to when we want to write 591219 in base 3, so there must be one triple satisfying this condition because we cannot write one number in base 10 in two different numbers in another base.

Therefore, $591219=1010001000000_{3}$ $(x,y,z)=(12,10,6)$

The sum is 28