Base of number base

$81x + 9y + z = 36z + 6y + x\\ \Rightarrow 80x + 3y = 35z\\ \Rightarrow 35 | (80x + 3y)\\ \Rightarrow 5 | (80x + 3y)$

(where $a | b$ means that $b$ is divisible by $a$ ).

Since $5 | 80x$ , it must be the case that $5 | 3y$ , so $5 | y$ . Since $0 \leq y \leq 9$ , we have $y = 0$ or $y = 5$ .

If $y = 0$ :

$35 | 80x\\ \Rightarrow 7 | 80x\\ \Rightarrow 7 | x\\ \Rightarrow x = 7$

But this would mean $z = 16$ , which is impossible since it must be a single digit.

If $y = 5$ :

$35 | (80x + 15)\\ \Rightarrow 7 | (80x + 15)\\ \Rightarrow 80x + 15 = 0 \pmod{7}\\ \Rightarrow 80 \pmod{7} \cdot x + 15 \pmod{7} = 0\\ \Rightarrow 3x + 1 = 0 \pmod{7}\\ \Rightarrow x = 2$

Substituting into either of the original equations gives us $z = 5$ , and indeed we can verify that $255_9 = 552_6$ . Hence $x + y + z = 12$ .

81x+9y+z=36z+6y+x . and since these numbers are used in the base 6 number and the base 9 number, 0<= x,y,z <= 5 . 80x+3y-35z=0 . x cant be zero because that would make xyz a two digit number. and z cant be zero because zyx would be a two digit number. if x=1, 80+3y-35z=0 . 3y=35z-80 . since 3y>=0, 35z>=80 . so z>=3 and 35z-80>=25 . but then 3y>=25 so y>=9 but y has to be less than or equal to 5, so that doesnt work, therefore x isnt equal to 1 . if x=2, 160+3y-35z=0 . 3y=35z-160 . again, since 3y>=0, 35z>=160 . so z>= 5. The only z that works which is greater than or equal to 5 is 5 itself, since each digit had to be between 0 and 5 inclusive. this means that z=5 . so 3y=35*5-160 . 3y=175-160 . 3y=15 . y=5 . therefore if x=2, y=5, and z=5 . you can try the rest for when x=3,4, and 5 and see they don't work. so the answer is 2+5+5=12 .

81x+9y+1z=36z+6y+1x

80x+3y=35z

X=3 Y=5 Z=5