2024 AIME II Problems/Problem 11

Problem

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

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

solution 2

$a^2(b+c)+b^2(a+c)+c^2(a+b) = 6000000$ , thus $a^2(300-a)+b^2(300-b)+c^2(300-c) = 6000000$ . Complete the cube to get $-(a-100)^3-(b-100)^3+(c-100)^3 = 9000000-30000(a+b+c)$ , which so happens to be 0. Then we have $(a-100)^3+(b-100)^3+(c-100)^3 = 0$ . We can use Fermat's last theorem here to note that one of a, b, c has to be 100. We have 200+200+200+1 = 601.

2024 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 10	Followed by Problem 12
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

Page

Toolbox

Search

2024 AIME II Problems/Problem 11

Contents

Problem

solution 1

solution 2

See also