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Let $N =\overline{abc}$ be a three-digit integer with distinct digits $a$, $b$, and $c$. What is the largest possible integer $N$ such that, when $N$ is cubed, the resulting integer ends with the same three digits as $N$?

Here is what I did: I know that $N^3\equiv N \pmod{1000}.$ That means that $N^3-N\equiv 0 \pmod{1000}$ or $N(N-1)(N+1)\equiv0 \pmod{1000}.$ However, I don't know how to quickly find numbers that fit the properties without brute force. What do I do?

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    $\begingroup$ surely either $N\equiv1($mod$ 1000)$ or $N\equiv999($mod$ 1000)$? $\endgroup$ – Seth Mar 13 at 2:58
  • $\begingroup$ There are other solutions to $N^3\equiv N\pmod {1000}$ besides $0, 1, $ and $999$ (e.g., 125), but $999$ works for this problem $\endgroup$ – J. W. Tanner Mar 13 at 3:08
  • $\begingroup$ However, $N$ has distinct digits, so 999 is not an answer. $\endgroup$ – A R Mar 13 at 3:12
  • $\begingroup$ Oops, I missed that the digits are distinct. Maybe $875?$ $\endgroup$ – J. W. Tanner Mar 13 at 3:18
  • $\begingroup$ I think you meant $N^\mathbb 3-N$ $\endgroup$ – J. W. Tanner Mar 13 at 15:41
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As stated in the question, we are looking for a solution of $N^3\equiv N\pmod{1000}.$

This means $1000$ divides $N^3-N$, so $125$ and $8$ divide $N^3-N=N(N-1)(N+1).$

$5$ can divide only one of $N, N-1$, and $N+1$, so this means $125$ divides $N$, $N-1$, or $N+1$.

That means, if $N$ is a positive integer less than $1000,$

$ N \in \{0,1,124,125,126,249,250,251,374,375,376,499,500,501,624,625,626,749,750,751,874,875,876,999\}.$

If $8$ divides $N(N-1)(N+1)$ then $N$ is odd or $8$ divides $N.$

That means $N\in\{0,1,125,249,251,375,376,499,501,624,625,749,751,875,999\}.$

Now that we have found solutions of $N^3\equiv N\pmod{1000}$, the problem is easily solved.

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OK, I got it. So, based on my initial trial, I know that $N(N-1)(N+1)\equiv0 \pmod{1000}.$ Now there are two cases. Why? Because $N+1$ and $N-1$ must have the same parity, meaning that they must both be even or they must both be odd. Note that $1000=2^3\times5^3$. Using this, we can rule out the case where $N$ is even. Why? Because if $N$ is even, $N+1$ and $N-1$ must be odd. $N$ must be divisible by 2, 4, or 8, but the two odd numbers should be divisible by 5, an impossibility (Note that $500$ is both divisible by 125 and 4, but not, 8, and since the other two numbers are odd. This means that $N$ must be odd, and if $N$ should be maximized, it should be a factor of 125. Note that there is at least one multiple of 4 next to a number ending in 25 or 75, so this assumption works. This gives us the largest three digit multiple of 125, $\boxed{875}$.

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    $\begingroup$ Why can't we have $N=376$ (even and $125$ divides $N-1$) as a solution to $N(N-1)(N+1)\equiv0\pmod{1000}$? $\endgroup$ – J. W. Tanner Mar 13 at 14:22

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