Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎solution 1)
Line 15: Line 15:
  
 
<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 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 <math>a+b+c = 300</math>.
 +
 +
Therefore,
 +
\[
 +
\left( a - 100 \right) \left( b - 100 \right) \left( c - 100 \right) = 0 .
 +
\]
 +
 +
Case 1: Exactly one out of <math>a - 100</math>, <math>b - 100</math>, <math>c - 100</math> 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 <math>a - 100 = 0</math> in Step 1. In this step, we determine <math>b</math> and <math>c</math>.
 +
 +
Recall <math>a + b + c = 300</math>. Thus, <math>b + c = 200</math>.
 +
Because <math>b</math> and <math>c</math> are nonnegative integers and <math>b - 100 \neq 0</math> and <math>c - 100 \neq 0</math>, the number of solutions is 200.
 +
 +
Following from the rule of product, the number of solutions in this case is <math>3 \cdot 200 = 600</math>.
 +
 +
Case 2: At least two out of <math>a - 100</math>, <math>b - 100</math>, <math>c - 100</math> are equal to 0.
 +
 +
Because <math>a + b + c = 300</math>, we must have <math>a = b = c = 100</math>.
 +
 +
Therefore, the number of solutions in this case is 1.
 +
 +
Putting all cases together, the total number of solutions is <math>600 + 1 = \boxed{\textbf{(601) }}</math>.
 +
 +
 +
 +
~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==
 
==See also==

Revision as of 16:59, 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)

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

[[Category:]] The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2024_AIME_II_Problems/Problem_11&oldid=215111"