The natural number $y$ is obtained from the number $x$ by rearranging it's digits. Suppose $x+y=10^{200}$. Prove that $x$ is divisible by 10. This is my question.I found that the sum of the digits of $x$+ the sum of the digits of $y$ is even.
$x+y=10^{200}$. Prove that $x$ is divisible by 10.
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algebra-precalculus
number-theory
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1Hint: If the natural number $y$ is obtained from the number $x$ by rearranging it's digits, then it must also be true that the natural number $x$ is obtained from the number $y$ by rearranging it's digits. – 2017-01-19
2 Answers
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Asume the last digit is $>0$. Let $x'=x-1$ be the number obtained from $x$ by decreasing the last digit. Then $x'+y$ consists of two-hundred nines. This means that in each place, $x'$ and $y$ have digits summing up to $9$ and never producing a carry. Hence the digit sum of $x'$ plus digit sum of $y$ is $200\cdot 9$. Then the original digit sum is $200\cdot 9+1$. This is odd, wheras you already found out that t should be even.
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Suppose contrary that $x$ is not divisible by $10$. If $x=\overline{x_{199}x_{198}\ldots x_1x_0}$ and $y=\overline{y_{199}y_{198}\ldots y_1y_0}$ are decimal representations of $x$ and $y$, then $x_0\neq 0$. Hence, we have $x_0+y_0=10$, whence $x_1+y_1=9$, $x_2+y_2=9$, $\ldots$, $x_{199}+y_{199}=9$. Do you see a contradiction now?