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What is the smallest positive integer $n$, such that there exist positive integers $a$ and $b$, with $b$ obtained from $a$ by a rearrangement of its digits, so that $a – b = 11\dots 1$ (The number of '$1$'s equal to $n$)?

Is there any example that can satisfy the equation?

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Hint 1. Since $b$ is obtained as a rearrangement of the digits of $a$, it follows that $a$ and $b$ have the same remainder when they are divided by $9$. Hence $a-b$ should be divisible by $9$.

Hint 2. Note that $$90-09=81,\ 190-019=171,\ 1290-0129=1161,\ 12390-01239=11151, \dots$$

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Since $b = \text{DigitsRearranged}(a) \rightarrow$ $a$ and $b$ have the same remainder when divided by $9$ $\rightarrow$ $a-b$ is divisible by $9$ $\rightarrow$ $n \ge 9$

And this is sufficient, because there is a solution for this $n$: $$a = 812,345,709 \\ b = 701,234,598 \\ a-b = 111,111,111$$

So the smallest $n$ is: $n = 9$.

This took a fair bit a fiddling around to find, toguh.

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