Difference between revisions of "2024 AIME II Problems/Problem 11"
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\end{equation*} | \end{equation*} | ||
| + | ==solution 1== | ||
| + | <math>ab(a+b)+bc(b+c)+ac(a+c)=300(ab+bc+ac)-3abc=6000000, 100(ab+bc+ac)-abc=2000000</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 | ||
==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 12:00, 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
Note 
. Thus, 
. There are 
cases for each but we need to subtract 
for 
. The answer is 
~Bluesoul
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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