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📅 May 4, 2026 👤 scraface

May 04, 2026

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\sum_{k=1}^{n} k^3 = \left( \frac{n(n+1)}{2} \right)^2
For $ n = 10 $:
\sum_{k=1}^{10} k^3 = \left( \frac{10 \cdot 11}{2} \right)^2 = (55)^2...
Now compute $ 3025 \mod 7 $. First, reduce 55 modulo 7:
\div 7 = 7 \text{ remainder } 6 \Rightarrow 55 \equiv 6 \pmod{7}
\Rightarrow 55^2 \equiv 6^2 = 36 \equiv 1 \pmod{7}
Alternatively, $ 3025 \div 7 = 432 \text{ remainder } 1 $, so $ 3025 \...
Therefore, the remainder is $ \boxed{1} $.
Question: What is the greatest common divisor of $ 2^{12} - 1 $ and $...
Solution: We use the identity:

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