If $x+y=10^{200}$ then prove that 50 divides $x$ Let $x$ be a positive integer and $y$ is another integer obtained after rearranging the digits of $x$. If $x+y=10^{200}$ then prove that $x$ is divisible by 50.
My attempt
Since $y$ is the digit rearrangement of $x$ so $x$ $\cong$ $y$ $\bmod{9}$ from there we get $x$ $\cong$ $5$ $\bmod{9}$ and $y$ $\cong$ $5$ $\bmod{9}$. Also possible last digits of $x$ and $y$ are
$(0,0), (1,9),(2,8),(3,7) , (4,6),(5,5),(6,4),(7,3),(8,2),(9,1)$. 
For last digits $(0,0), (2,8), (4,6),(6,4),(8,2)$ divisibility by $2$ is ensured but divisibility by $25$ and general case is eluding me. Can someone help? Thanks in advance.

We can assume $x$ and $y$ are non-zero. So, with suitable initial $0$-padding, each has $200$ digits.
If $x$ ends in $00$ we are finished. Suppose now that $x$ ends in $0$ but not $00$. Then the next to last digits of $x$ and $y$ are $10$'s complements of each other, and non-zero. Each of $x$ and $y$ has $198$ digits that are $9$'s complements of each other. These come in pairs, since the digits are permuted. So the next to last digit of $x$ and $y$ are equal, and therefore each is $5$.
And we cannot have last digit non-zero, for in the rest of $x$ and $y$ the digits come in $9$'s complement pairs, and $199$ is odd.
