Who can send solution of this problem

Chevrolet23 wrote: Who can send solution of this problem İ dont have any idea

Let $x=x_1x_2x_3x_4x_5x_6$ where $x_i$ are digits We have $x_6+y_6=10$ $x_5+y_5 \geq 10$ because if not than $z_5=x_4+y_4$ is even $x_4+y_4 \geq 10$ $x_3+y_3 \geq 10$ $x_2+y_2 \geq 10$ $x_1+y_1 \geq 10$ $x_i+y_i=10$ have $5$ solutions and $x_i+y_i \geq 10$ has $15$ solutions So total we have $5*15^5$ solutions.

RagvaloD wrote: Let $x=x_1x_2x_3x_4x_5x_6$ where $x_i$ are digits We have $x_6+y_6=10$ $x_5+y_5 \geq 10$ because if not than $z_5=x_4+y_4$ is even $x_4+y_4 \geq 10$ $x_3+y_3 \geq 10$ $x_2+y_2 \geq 10$ $x_1+y_1 \geq 10$ $x_i+y_i=10$ have $5$ solutions and $x_i+y_i \geq 10$ has $15$ solutions So total we have $5*15^5$ solutions. Hi,good solution but i think that x1+y1 can take any value,what about you?

Telman wrote: RagvaloD wrote: Let $x=x_1x_2x_3x_4x_5x_6$ where $x_i$ are digits We have $x_6+y_6=10$ $x_5+y_5 \geq 10$ because if not than $z_5=x_4+y_4$ is even $x_4+y_4 \geq 10$ $x_3+y_3 \geq 10$ $x_2+y_2 \geq 10$ $x_1+y_1 \geq 10$ $x_i+y_i=10$ have $5$ solutions and $x_i+y_i \geq 10$ has $15$ solutions So total we have $5*15^5$ solutions. Hi,good solution but i think that x1+y1 can take any value,what about you? Hi, if $x_1+y_1$ less than 10, x+y can't be 7 digit number but 10z is 7 digit number