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Home » Teemo Counter Explained: Why This Simple Trick Dominated the Meta Forever! - Baxtercollege

Teemo Counter Explained: Why This Simple Trick Dominated the Meta Forever! - Baxtercollege

📅 March 10, 2026 👤 scraface

Mar 10, 2026

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📌 Solution: By De Moivre's Theorem:

📌 oxed{i}Question: What is the smallest positive integer whose cube ends in the digits $888$?

📌 Solution: We seek the smallest positive integer $ n $ such that $ n^3 \equiv 888 \pmod{1000} $. 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.

📌 First, modulo $8$:

📌 $ n^3 \equiv 888 \pmod{8} $. Since $888 \div 8 = 111$, $888 \equiv 0 \pmod{8}$.

📌 So $ n^3 \equiv 0 \pmod{8} $, which implies $ n \equiv 0 \pmod{2} $, and in fact $ n $ must be divisible by $2$. But to get $n^3 \equiv 0 \pmod{8}$, $n$ must be divisible by $2$, and checking small cubes mod $8$: $0^3 = 0$, $2^3 = 8 \equiv 0$, $4^3 = 64 \equiv 0$, $6^3 = 216 \equiv 0$. So all even $n$ satisfy $n^3 \equiv 0 \pmod{8}$. So $ n \equiv 0 \pmod{2} $ is sufficient,

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