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Let the integer $y$ be obtained by rearranging the digits of the integer $x$. If $x+y = 10^{200}$, prove that $x$ is divisible by $10$.

Let $x = \overline{a_0 \ldots a_n},$ so that $y = \overline{a_n \ldots a_0}$. How can we use the fact that we are rearranging the digits of $x$?

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Assume otherwise, so that $x = \overline{a_1 \ldots a_{200}}$ has $a_{200} \neq 0$. Let $y = \overline{b_1 \ldots b_{200}}$. Then we must have $$x+y = \overline{a_1 \ldots a_{200}}+\overline{b_1 \ldots b_{200}} = 10^{200}$$ and so $a_{200}+b_{200} = 10, a_{199}+b_{199} = 9, a_{198}+b_{198} = 9, \ldots, a_1+b_1 = 9$. Adding up all of the equations gives $$(a_1+\cdots+a_{200})+(b_1+\cdots+b_{200}) = 10+199 \cdot 9.$$ The left-hand side is even, while the right-hand side is odd, contradiction.

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    ... which strongly suggests that $x=y=5\cdot 10^{199}$, the "rearrangement" being a shuffle of the zeroes, perhaps. – 2017-01-16