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 11: Line 11:
  
 
~Bluesoul
 
~Bluesoul
 +
 +
==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.
 +
 
==See also==
 
==See also==
 
{{AIME box|year=2024|num-b=10|num-a=12|n=II}}
 
{{AIME box|year=2024|num-b=10|num-a=12|n=II}}

Revision as of 15:32, 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.

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=215099"