A Question for Mr. Modulator

The title suggests we want to use modular arithmetic; in fact, Fermat's little theorem , $a^{p-1} \equiv 1 \pmod{p}$ (where $p$ is a prime and $a$ is an integer not divisible by $p$ ) almost works straight away. The problem is of course that $p-1$ can't be $5$ ; the next best thing is to try $p=2\cdot 5+1=11$ .

Fermat's little theorem tells us that $a^{10}$ is congruent to either $0$ or $1$ mod $11$ ; therefore $a^5$ is congruent to one of $\{-1,0,1\}$ . By considering all combinations of these, we find that $x^5-y^5-z^5$ has to be congruent to one of $\{0, 1, 2, 3, 8, 9, 10\}$

Among the options, we find (see below for a non-calculator method) that $4217000\equiv 7 \pmod{11}$ is the only one that doesn't belong to this set, so the answer (without explicitly finding $(x,y,z)$ for the others) is $\boxed{4217000}$ .

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To find $n \pmod{11}$ manually, we can just use long division, but this is a little tedious for very large $n$ (and also, we're interested in the remainder, not the quotient).

Instead, say $n$ has $k$ digits, and $n=\overline{d_{k-1} d_{k-2} \cdots d_0}$ . Since $10\equiv-1 \pmod{11}$ , we have $\begin{aligned} \overline{d_{k-1} d_{k-2} \cdots d_0} &\equiv \sum_{r=0}^{k-1} 10^r d_r \\ &\equiv \sum_{r=0}^{k-1} (-1)^r d_r \pmod{11} \end{aligned}$

ie we need to find the sums of the digits in odd and even positions (counting from the right).

For example, $4217000 \equiv (4+1+0+0)-(2+7+0) = -4 \equiv 7 \pmod{11}$