10. What is the smallest integer $x$ larger than 1 such that $x^{2}$ ends in the same three digits as $x$ does?
Solution: 376
The condition is that $1000 \mid x^{2}-x=x(x-1)$. Since $1000=2^{3} \cdot 5^{3}$, and 2 cannot divide both $x$ and $x-1,2^{3}=8$ must divide one of them. Similarly, $5^{3}=125$ must divide either $x$ or $x-1$. We try successive values of $x$ that are congruent to 0 or 1 modulo 125 and see which ones have the property that $x$ or $x-1$ is divisible by 8 . It is easy to check that $125,126,250,251$, and 375 do not work, but the next value, 376 , does, so this is the answer.