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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 \( \boxed{50} \). 📰 Frage: Wie viele der 150 kleinsten positiven ganzen Zahlen sind kongruent zu 3 (mod 7)? 📰 Lösung: Wir suchen die Anzahl der positiven ganzen Zahlen \( n \le 150 \), sodass \( n \equiv 3 \pmod{7} \). Solche Zahlen haben die Form \( n = 7k + 3 \). Wir benötigen \( 7k + 3 \le 150 \), also \( 7k \le 147 \) → \( k \le 21 \). Da \( k \ge 0 \), reichen \( k = 0, 1, 2, \dots, 21 \), also insgesamt 22 Werte. Somit gibt es \( \boxed{22} \) solche Zahlen. 📰 Vikings Head Coach 263236 📰 Tagline Frenzy Watch Uipath Stock Jump To New Heights In 2024 Dont Miss Out 7870126 📰 Zodiac For October 23Rd 7719815 📰 How To Convert Heic To Jpg On Mac 904450 📰 Touchdown Wings Just Stole My Heart Againwatch What Happens Next 9621831 📰 Top Rated Apple Phones 4864506 📰 Adding A Tick In Word 5029561 📰 Why Top Developers Join The Oracle Cloud Communityyou Wont Want To Miss This 7988717 📰 Best Day To Buy Airline Tickets 8943459 📰 Jac Schaeffer 8634847 📰 Anime Boy Drawing Goes Viral Join Millions In Completely Losing It 5998831 📰 Long Branch Hotels 1203933 📰 Zoho Workdrive Review The Ultimate File Storage Youve Been Searching For 317044 📰 Wells Fargo Active Cash Balance Transfer 9854119 📰 Bellmore New York 7728293