Solution: We seek the smallest positive integer $ n $ such that $ n^3 \equiv 888 \pmod1000 $. This means we want the last three digits of $ n^3 $ to be $888$. We solve this congruence modulo $1000$. Since $1000 = 8 imes 125$, and $8$ and $125$ are coprime, we can use the Chinese Remainder Theorem by solving modulo $8$ and modulo $125$ separately. - Nelissen Grade advocaten