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

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==Problem==
 
==Problem==
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Find the number of triples of nonnegative integers \((a,b,c)\) satisfying \(a + b + c = 300\) and
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Find the number of triples of nonnegative integers <math>(a,b,c)</math> satisfying <math>a + b + c = 300</math> and
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\begin{equation*}
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<cmath>a^2b + a^2c + b^2a + b^2c + c^2a + c^2b = 6,000,000.</cmath>
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a^2b + a^2c + b^2a + b^2c + c^2a + c^2b = 6,000,000.
 
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\end{equation*}
 
  
 
==Solution 1==
 
==Solution 1==
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Note <math>(100-a)(100-b)(100-c)=1000000-10000(a+b+c)+100(ab+bc+ac)-abc=0</math>. Thus, <math>a/b/c=100</math>. There are <math>201</math> cases for each but we need to subtract <math>2</math> for <math>(100,100,100)</math>. The answer is <math>\boxed{601}</math>
 
Note <math>(100-a)(100-b)(100-c)=1000000-10000(a+b+c)+100(ab+bc+ac)-abc=0</math>. Thus, <math>a/b/c=100</math>. There are <math>201</math> cases for each but we need to subtract <math>2</math> for <math>(100,100,100)</math>. The answer is <math>\boxed{601}</math>
  
−
~Bluesoul
+
~Bluesoul,Shen Kislay Kai
  
 
==Solution 2==  
 
==Solution 2==  
  
−
<math>a^2(b+c)+b^2(a+c)+c^2(a+b) = 6000000</math>, thus <math>a^2(300-a)+b^2(300-b)+c^2(300-c) = 6000000</math>. Complete the cube to get <math>-(a-100)^3-(b-100)^3+(c-100)^3 = 9000000-30000(a+b+c)</math>, which so happens to be 0. Then we have <math>(a-100)^3+(b-100)^3+(c-100)^3 = 0</math>. 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.
+
<math>a^2(b+c)+b^2(a+c)+c^2(a+b) = 6000000</math>, thus <math>a^2(300-a)+b^2(300-b)+c^2(300-c) = 6000000</math>. Complete the cube to get <math>-(a-100)^3-(b-100)^3+(c-100)^3 = 9000000-30000(a+b+c)</math>, which so happens to be 0. Then we have <math>(a-100)^3+(b-100)^3+(c-100)^3 = 0</math>. We can use Fermat's last theorem here to note that one of <math>a, b, c</math> has to be 100. We have <math>200+200+200+1 = 601.</math>
  
 
==Solution 3==
 
==Solution 3==
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Therefore,
 
Therefore,
−
\[
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<cmath>
 
\left( a - 100 \right) \left( b - 100 \right) \left( c - 100 \right) = 0 .
 
\left( a - 100 \right) \left( b - 100 \right) \left( c - 100 \right) = 0 .
−
\]
+
</cmath>
  
 
Case 1: Exactly one out of <math>a - 100</math>, <math>b - 100</math>, <math>c - 100</math> is equal to 0.
 
Case 1: Exactly one out of <math>a - 100</math>, <math>b - 100</math>, <math>c - 100</math> is equal to 0.
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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> satisfies the conditions.
 
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> satisfies the conditions.
  
−
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.
+
Now to count the possible solutions, we note that when <math>x \ne 0</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>1</math> to <math>100</math>, there are <math>100 \cdot 3! = 600</math> triples in this case. When <math>x = 0</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.
 
In total, there are thus <math>\boxed{601}</math> distinct triples.
  
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~GaloisTorrent <3
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~GaloisTorrent <Shen Kislay Kai>
 +
 
 +
- minor edit made by MEPSPSPSOEODODODO
  
 
==Solution 5==
 
==Solution 5==

Revision as of 23:08, 5 September 2024

Problem

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

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,Shen Kislay Kai

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 $100$ satisfies the conditions.

Now to count the possible solutions, we note that when $x \ne 0$, the three roots are distinct; thus there are $3! = 6$ ways to order the three roots. As we can choose $x$ from $1$ to $100$, there are $100 \cdot 3! = 600$ triples in this case. When $x = 0$, 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 <Shen Kislay Kai>

- minor edit made by MEPSPSPSOEODODODO

Solution 5

Let's define $a=100+x$, $b=100+y$, $c=100+z$. Then we have $x+y+z=0$ and $6000000 = \sum a^2(b+c)$

$= \sum (100+x)^2(200-x) = \sum (10000+200x+x^2)(200-x) = \sum (20000 - 10000 x + x(40000-x^2))$

$= \sum (20000 + 30000 x -x^3) = 6000000 - \sum x^3$, so we get $x^3 + y^3 + z^3 = 0$. Then from $x+y+z = 0$, we can find $0 = x^3+y^3+z^3 = x^3+y^3-(x+y)^3 = 3xyz$, which means that one of $a$, $b$,$c$ must be 0. There are 201 solutions for each of $a=0$, $b=0$ and $c=0$, and subtract the overcounting of 2 for solution $(200, 200, 200)$, the final result is $201 \times 3 - 2 = \boxed{601}$.

Dan Li


dan

Video Solution

https://youtu.be/YMYe9chPLdY

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

See also

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

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