2024 AIME II Problems/Problem 11

Problem

Find the number of triples of nonnegative integers $(a,b,c)$ satisfying $a + b + c = 300$ and \begin{equation*} a^2b + a^2c + b^2a + b^2c + c^2a + c^2b = 6,000,000. \end{equation*}

solution 1

$ab(a+b)+bc(b+c)+ac(a+c)=300(ab+bc+ac)-3abc=6000000, 100(ab+bc+ac)-abc=2000000$

Note $(100-a)(100-b)(100-c)=1000000-10000(a+b+c)+100(ab+bc+ac)-abc=0$ . Thus, $a/b/c=100$ . There are $201$ cases for each but we need to subtract $2$ for $(100,100,100)$ . The answer is $\boxed{601}$

~Bluesoul

2024 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 10	Followed by Problem 12
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2024 AIME II Problems/Problem 11

Problem

solution 1

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