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Home » An environmental project uses $120,000 to fund 3 tasks: sensors, delivery, and training. Sensors cost twice delivery, and training costs $10,000 less than delivery. How much goes to sensors? - AMAZONAWS

An environmental project uses $120,000 to fund 3 tasks: sensors, delivery, and training. Sensors cost twice delivery, and training costs $10,000 less than delivery. How much goes to sensors? - AMAZONAWS

How $120,000 Was Allocated to a Green Initiative — Sensors Got $60,000

📅 March 11, 2026 👤 scraface
Mar 11, 2026
An environmental project uses $120,000 to fund 3 tasks: sensors, delivery, and training. Sensors cost twice delivery, and training costs $10,000 less than delivery. How much goes to sensors?

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📌 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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