2024 AIME II Problems/Problem 11

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Problem

Find the number of triples of nonnegative integers (a,b,c) satisfying a+b+c=300 and a2b+a2c+b2a+b2c+c2a+c2b=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.

Solution 3

We have a2b+a2c+b2a+b2c+c2a+c2b=ab(a+b)+bc(b+c)+ca(c+a)=ab(300c)+bc(300a)+ca(300b)=300(ab+bc+ca)3abc=3((a100)(b100)(c100)104(a+b+c)+106)=3((a100)(b100)(c100)2106)=6106. The first and the fifth equalities follow from the condition that $a+b+c = 300$.

Therefore, \[ \left( a - 100 \right) \left( b - 100 \right) \left( c - 100 \right) = 0 . \]

Case 1: Exactly one out of $a - 100$, $b - 100$, $c - 100$ is equal to 0.

Step 1: We choose which term is equal to 0. The number ways is 3.

Step 2: For the other two terms that are not 0, we count the number of feasible solutions.

W.L.O.G, we assume we choose $a - 100 = 0$ in Step 1. In this step, we determine $b$ and $c$.

Recall $a + b + c = 300$. Thus, $b + c = 200$. Because $b$ and $c$ are nonnegative integers and $b - 100 \neq 0$ and $c - 100 \neq 0$, the number of solutions is 200.

Following from the rule of product, the number of solutions in this case is $3 \cdot 200 = 600$.

Case 2: At least two out of $a - 100$, $b - 100$, $c - 100$ are equal to 0.

Because $a + b + c = 300$, we must have $a = b = c = 100$.

Therefore, the number of solutions in this case is 1.

Putting all cases together, the total number of solutions is $600 + 1 = \boxed{\textbf{(601) }}$.


~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Video Solution

https://youtu.be/YMYe9chPLdY

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

See also

2024 AIME II (ProblemsAnswer KeyResources)
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

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