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Home » Solving for \(n\): - Aurero

Solving for \(n\): - Aurero

📅 March 13, 2026 👤 scraface

Mar 13, 2026

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📌 n^2 = 100k^2 + 120k + 36 \equiv 20k + 36 \pmod{100}

📌 Set \( 20k + 36 \equiv 76 \pmod{100} \):

📌 20k \equiv 40 \pmod{100} \implies k \equiv 2 \pmod{5}

📌 So \( k = 5m + 2 \), \( n = 10(5m + 2) + 6 = 50m + 26 \). Smallest \( n = 26 \). Check: \( 26^2 = 676 \), ends in 76 — valid.

📌 Between 24 and 26, the smaller is 24. Therefore, the smallest positive integer whose square ends in 76 is \( \boxed{24} \).

📌 Question:** What is the remainder when the sum \( 1^3 + 2^3 + 3^3 + \dots + 12^3 \) is divided by 9?

📌 For \( n = 12 \):

📌 \sum_{k=1}^{12} k^3 = \left( \frac{12 \cdot 13}{2} \right)^2 = (78)^2 = 6084

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