Question

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

Answer

Solution:

The left side is the symmetric sum $(a + b + c)(ab + bc + ca) - 3abc = 300q - 3p,$ where $q = ab + bc + ca$ and $p = abc.$ So the condition is $100q - p = 2{,}000{,}000.$ Now expand $(100 - a)(100 - b)(100 - c) = 10^6 - 10^4 (a + b + c) + 100q - p = (100q - p) - 2 \cdot 10^6,$ using $a + b + c = 300.$ The condition holds exactly when this product is $0,$ that is, when at least one of $a, b, c$ equals $100.$

If $a = 100,$ then $b + c = 200,$ giving $201$ triples, and likewise for $b$ and $c:$ $3 \cdot 201 = 603.$ A triple counted more than once has two variables equal to $100,$ which forces the third to be $100$ as well; the triple $(100, 100, 100)$ is counted three times, so the total is $603 - 2 = 601.$

2024 AIME II Problem 11

Solution:

Problem 11 in Other Years