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Home » Ready for a Whimsical Round? Peter Pan Putt-Putt Golf in Austin Will Blow Your Mind! - Network4

Ready for a Whimsical Round? Peter Pan Putt-Putt Golf in Austin Will Blow Your Mind! - Network4

📅 May 19, 2026 👤 scraface

May 19, 2026

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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:
\gcd(2^a - 1, 2^b - 1) = 2^{\gcd(a,b)} - 1

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