Difference between revisions of "2024 AIME II Problems/Problem 11"

Revision as of 20:12, 9 February 2024

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

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 \begin{align*} & a^2 b + a^2 c + b^2 a + b^2 c + c^2 a + c^2 b \\ & = ab \left( a + b \right) + bc \left( b + c \right) + ca \left( c + a \right) \\ & = ab \left( 300 - c \right) + bc \left( 300 - a \right) + ca \left( 300 - b \right) \\ & = 300 \left( ab + bc + ca \right) - 3 abc \\ & = -3 \left( \left( a - 100 \right) \left( b - 100 \right) \left( c - 100 \right) - 10^4 \left( a + b + c \right) + 10^6 \right) \\ & = -3 \left( \left( a - 100 \right) \left( b - 100 \right) \left( c - 100 \right) - 2 \cdot 10^6 \right) \\ & = 6 \cdot 10^6 . \end{align*} 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)

Solution 4

We will use Vieta's formulas to solve this problem. We assume $a + b + c = 300$ , $ab + bc + ca = m$ , and $abc = n$ . Thus $a$ , $b$ , $c$ are the three roots of a cubic polynomial $f(x)$ .

We note that $300m = (a + b + c)(ab + bc + ca)=\sum_{cyc} a^2b + 3abc = 6000000 + 3n$ , which simplifies to $100m - 2000000 = n$ .

Our polynomial $f(x)$ is therefore equal to $x^3 - 300x^2 + mx - (100m - 2000000)$ . Note that $f(100) = 0$ , and by polynomial division we obtain $f(x) = (x - 100)(x^2 - 200x - (m-20000))$ .

We now notice that the solutions to the quadratic equation above are $x = 100 \pm \frac{\sqrt{200^2 - 4(m - 20000)}}{2} = 100 \pm \sqrt{90000 - 4m}$ , and that by changing the value of $m$ we can let the roots of the equation be any pair of two integers which sum to $200$ . Thus any triple in the form $(100, 100 - x, 100 + x)$ where $x$ is an integer between $0$ and $100$ .

Now to count the possible solutions, we note that when $x \ne 100$ , the three roots are distinct; thus there are $3! = 6$ ways to order the three roots. As we can choose $x$ from $0$ to $99$ , there are $100 \cdot 3! = 600$ triples in this case. When $x = 100$ , all three roots are equal to $100$ , and there is only one triple in this case.

In total, there are thus $\boxed{601}$ distinct triples.

~GaloisTorrent <3

Video Solution

https://youtu.be/YMYe9chPLdY

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

@@ Line 65: / Line 65: @@
 ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
+==Solution 4==
+We will use Vieta's formulas to solve this problem. We assume <math>a + b + c = 300</math>, <math>ab + bc + ca = m</math>, and <math>abc = n</math>. Thus <math>a</math>, <math>b</math>, <math>c</math> are the three roots of a cubic polynomial <math>f(x)</math>.
+We note that <math>300m = (a + b + c)(ab + bc + ca)=\sum_{cyc} a^2b + 3abc = 6000000 + 3n</math>, which simplifies to <math>100m - 2000000 = n</math>.
+Our polynomial <math>f(x)</math> is therefore equal to <math>x^3 - 300x^2 + mx - (100m - 2000000)</math>. Note that <math>f(100) = 0</math>, and by polynomial division we obtain <math>f(x) = (x - 100)(x^2 - 200x - (m-20000))</math>.
+We now notice that the solutions to the quadratic equation above are <math>x = 100 \pm \frac{\sqrt{200^2 - 4(m - 20000)}}{2} = 100 \pm \sqrt{90000 - 4m}</math>, and that by changing the value of <math>m</math> we can let the roots of the equation be any pair of two integers which sum to <math>200</math>. Thus any triple in the form <math>(100, 100 - x, 100 + x)</math> where <math>x</math> is an integer between <math>0</math> and <math>100</math>.
+Now to count the possible solutions, we note that when <math>x \ne 100</math>, the three roots are distinct; thus there are <math>3! = 6</math> ways to order the three roots. As we can choose <math>x</math> from <math>0</math> to <math>99</math>, there are <math>100 \cdot 3! = 600</math> triples in this case. When <math>x = 100</math>, all three roots are equal to <math>100</math>, and there is only one triple in this case.
+In total, there are thus <math>\boxed{601}</math> distinct triples.
+~GaloisTorrent <3
 ==Video Solution==

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