If $a+b=10^{50}$ then show $10|a$

Assume $a \not \equiv 0 \pmod {10}$. Note that $a$ must be smaller than $10^{50}$, yet bigger than $10^{49}$.

Let $$a=\sum_{i=0}^{49}a_{i}10^{i}$$ Then let $(b_{0}, b_{1}, b_{2}, b_{3}, b_{4}, \dots, b_{49})$ be the digits of $b$ in order, and because of the definition of $b$ it follows that they are a rearrangement of $(a_{0}, a_{1}, a_{2}, a_{3}, a_{4}, \dots, a_{49})$.

So $$10^{50}=\sum_{i=i}^{49}(a_{i}+b_{i})10^{i}$$

Since $a_{0} \neq 0$, we have that $a_{0}+b_{0}=10$. Because of carrying, we can use this to claim that $a_{1}+b_{1}=9$, and so on. Eventually, we note that for $n \ge i \ge 1 $ we have $a_{i}+b_{i}=9$. So we have $$\sum_{i=i}^{49}(a_{i}+b_{i})=10 +9 +9+ \dots +9= 9 \times 49+10 =451$$ However, since $(b_{0}, b_{1}, b_{2}, b_{3}, b_{4}, \dots, b_{49})$ is a rearrangement of $(a_{0}, a_{1}, a_{2}, a_{3}, a_{4}, \dots, a_{49})$, we have $$451=\sum_{i=i}^{49}(a_{i}+b_{i})=2\sum_{i=0}^{49}a_{i}$$ A contradiction, since $\sum\limits_{i=0}^{49}a_{i}$ is a natural.