fprosk wrote: We have shortened the usual notation indicating with a sub-index the number of times that a digit is conseutively repeated. For example, $1119900009$ is denoted $1_3 9_2 0_4 9_1$. Find $(x, y, z)$ if $2_x 3_y 5_z + 3_z 5_x 2_y = 5_3 7_2 8_3 5_1 7_3$ Can anybody put this in LaTeX? Thanks. Just added dollar signs...

Oh wow thanks. I'm sorry but I really don't have much practive with LaTeX.

Well, just write the two summands one below each other and do addition digit per digit from right to left. Assume that at some point there is a carry to the next digit. Then the rightmost point where this happens must have been an addition of 5 and 5. But since there has not been a carry before the resulting digit will be 0 which does not occur in the result. Thus, there is no carry in this addition. Since a digit 7 at the end occurs iff we add 5 and 2 it follows that $3=\min(y,z)$ Also, easy to see that $x+y+z=12$ since both summands have $x+y+z$ digits and the result has 12 digits starting with a digit 5. But because there is no carry we also know that cross sum adds up so $2x+3y+5z+3z+5x+2y=15+14+24+5+21$ and hence $7x+5y+8z=79$ which gives $2x+3z=19$ and so either $z=3$ which implies $x=5$ and $y=4$, either $y=3$ which implies $x+z=9$ and hence z=1, x=8$ contradicting $\min(y,z)=3$. So, the only possibility is $x=5, y=4, z=3$ which is indeed a solution.