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📅 March 1, 2026 👤 scraface

Mar 01, 2026

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Solution: We compute the sum $ \sum_{k=1}^{24} 3^k \mod 7 $. This is a...
\sum_{k=1}^{24} 3^k = 3 \cdot \frac{3^{24} - 1}{3 - 1} = \frac{3}{2} (...
But working modulo $7$, we can compute the sum directly using cyclicit...
3^1 &\equiv 3 \mod 7 \\
3^2 &\equiv 9 \equiv 2 \mod 7 \\
3^3 &\equiv 3 \cdot 2 = 6 \mod 7 \\
3^4 &\equiv 3 \cdot 6 = 18 \equiv 4 \mod 7 \\
3^5 &\equiv 3 \cdot 4 = 12 \equiv 5 \mod 7 \\
3^6 &\equiv 3 \cdot 5 = 15 \equiv 1 \mod 7 \\
So the powers of $3 \mod 7$ repeat every 6 terms. The cycle length is...

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