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Home » How Top Players Are Exploiting Ingress MMO Mechanics Like a Pro! - 500apps

How Top Players Are Exploiting Ingress MMO Mechanics Like a Pro! - 500apps

📅 March 14, 2026 👤 scraface

Mar 14, 2026

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📌 = z^3 + \left(\frac{-407}{3}\right)z + \left(\frac{512 - 1536 + 648 - 486}{27}\right) = z^3 - \frac{407}{3}z - \frac{1422}{27} = z^3 - \frac{407}{3}z - \frac{474}{9}

📌 = z^3 - \frac{407}{3}z - \frac{158}{3}

📌 Diskriminante eines kubischen \( z^3 + pz + q \): \( \Delta = -4p^3 - 27q^2 \)

📌 Hier: \( p = -\frac{407}{3} \), \( q = -\frac{158}{3} \)

📌 \Delta = -4\left(-\frac{407}{3}\right)^3 - 27\left(-\frac{158}{3}\right)^2 = -4\left(-\frac{67432643}{27}\right) - 27\cdot\frac{24964}{9}

📌 Besser: positives Vorzeichen:

📌 \( -4p^3 = -4 \cdot \left( -\frac{407^3}{27} \right) = \frac{4}{27} \cdot 67432643 \), huge positive

📌 \( 27q^2 = 27 \cdot \frac{24964}{9} = 3 \cdot 24964 = 74892 \)

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Question: What is the remainder when the sum $ 1^3 + 2^3 + 3^3 + \dots...
Solution: We use the formula for the sum of cubes:
\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} $.

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