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Home » Ponchatoula: This Hidden Louisiana Gem Will Blow Your Mind! - AMAZONAWS

Ponchatoula: This Hidden Louisiana Gem Will Blow Your Mind! - AMAZONAWS

Ponchatoula: This Hidden Louisiana Gem Will Blow Your Mind!

📅 March 11, 2026 👤 scraface

Mar 11, 2026

Ponchatoula: This Hidden Louisiana Gem Will Blow Your Mind!

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📌 144y^2 + 3600 \equiv 0 \pmod{25} \Rightarrow 144y^2 \equiv -3600 \pmod{25}.

📌 25x^2 = 144(25k^2) + 3600 = 3600k^2 + 3600 = 3600(k^2 + 1).

📌 Lösung: Sei \( d = \gcd(a,b) \). Dann gilt \( a = d \cdot m \) und \( b = d \cdot n \), wobei \( m \) und \( n \) teilerfremde ganze Zahlen sind. Dann gilt \( a + b = d(m+n) = 100 \). Also muss \( d \) ein Teiler von 100 sein. Um \( d \) zu maximieren, minimieren wir \( m+n \), wobei \( m \) und \( n \) teilerfremd sind. Der kleinste mögliche Wert von \( m+n \) mit \( m,n \ge 1 \) und \( \gcd(m,n)=1 \) ist 2 (z. B. \( m=1, n=1 \)). Dann ist \( d = \frac{100}{2} = 50 \). Prüfen: \( a = 50, b = 50 \), \( \gcd(50,50) = 50 \), und \( a+b=100 \). Somit ist 50 erreichbar. Ist ein größerer Wert möglich? Wenn \( d > 50 \), dann \( d \ge 51 \), also \( m+n = \frac{100}{d} \le \frac{100}{51} < 2 \), also \( m+n < 2 \), was unmöglich ist, da \( m,n \ge 1 \). Daher ist der größtmögliche Wert \( \boxed{50} \).

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