Space Rockets

Space Rockets is a captivating and stunning space-themed arcade game. You will navigate through vast expanses and touch down on various intriguing planets for exploration.<|endoftext|>... I'm sorry, I do not understand what you are asking. Can you please provide more context or information?<|endoftext|><|endoftext|> # 2016 AMC 12A Problems/Problem 1. ## Contents. 1 Problem 2 Solution 3 Video Solution 4 See Also ## Problem. What is $24+48+72+...+2400$? $\textbf{(A)}\ 240100\qquad\textbf{(B)}\ 240200\qquad\textbf{(C)}\ 240300\qquad\textbf{(D)}\ 240400\qquad\textbf{(E)}\ 240500$ ## Solution. We can factor out $24$ from each term to get $24(1+2+3+...+100)$. The sum of the first $n$ positive integers is $\frac{n(n+1)}{2}$, so the sum of the first $100$ positive integers is $\frac{100(101)}{2}=5050$. Therefore, the sum is $24(5050)=\boxed{\textbf{(B)}\ 240200}$. ## Video Solution. https://youtu.be/8WrdYLw9_ns ~savannahsolver <|endoftext|>## Mathematical Forums ## Category: High School Olympiads ## Topic: Inequality ## Views: 338 ## [enter: math-user1, num_posts=697, num_likes_received=372] ## [math-user1, num_likes=1] Let $a,b,c$ be positive real numbers such that $a+b+c=3$. Prove that $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+3\geq 2(ab+bc+ca)$ ## [enter: math-user2, num_posts=467, num_likes_received=180] ## [math-user2, num_likes=0] By AM-GM, $ab+bc+ca\le\frac{\left(a+b+c\right)^2}{3}=3$. So we need to prove $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\ge3$. By AM-GM, $\frac{a}{b}+\frac{a}{b}+\frac{b}{c}\ge3\sqrt[3]{\frac{a^2}{bc}}=3\sqrt[3]{\frac{a^3}{abc}}=3\sqrt[3]{\frac{a^3}{abc}\cdot\frac{b^3}{abc}\cdot\frac{c^3}{abc}}=3$. Summing cyclically gives the desired result