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Home » Exploring *NO MAN’S SKY* Like Never Before: Mind-Blowing Secrets Unlocked! - Baxtercollege

Exploring NO MAN’S SKY Like Never Before: Mind-Blowing Secrets Unlocked! - Baxtercollege

📅 March 11, 2026 👤 scraface

Mar 11, 2026

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📌 An epidemiologist tracking a disease outbreak observes that the total number of cases over two consecutive weeks is 144, and the greatest common divisor of the weekly case counts is as large as possible. What is the largest possible value of \( \gcd(a, b) \) if \( a + b = 144 \) and \( a, b \) are positive integers?

📌 Let \( a + b = 144 \), and let \( d = \gcd(a, b) \). Then we can write \( a = d \cdot m \), \( b = d \cdot n \), where \( m \) and \( n \) are coprime positive integers. Then:

📌 a + b = d(m + n) = 144.

📌 Thus, \( d \) must be a divisor of 144, and \( m + n = \frac{144}{d} \). Since \( m \) and \( n \) are coprime positive integers, \( m + n \geq 2 \), so \( \frac{144}{d} \geq 2 \Rightarrow d \leq 72 \).

📌 To maximize \( d \), we take the largest divisor of 144 such that \( \frac{144}{d} \geq 2 \) and there exist coprime \( m, n \) with \( m + n = \frac{144}{d} \).

📌 The largest divisor of 144 is 144 itself, but then \( m + n = 1 \), impossible. Next, try \( d = 72 \): then \( m + n = 2 \), so \( m = n = 1 \), which are coprime. Then \( a = 72 \), \( b = 72 \), and \( \gcd(72, 72) = 72 \). Valid.

📌 So the maximum possible \( \gcd(a, b) \) is:

📌 \boxed{72}

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