2024 AIME II Problem 11

Question

Find the number of triples of nonnegative integers

(a, b, c)

satisfying

a + b + c = 300

and

\begin{aligned} &a^2 b + a^2 c \\ &\quad {}+ b^2 a + b^2 c \\ &\quad {}+ c^2 a + c^2 b = 6{,}000{,}000. \end{aligned}

Answer

With

a + b + c = 300,

the given sum equals

300(ab + bc + ca) - 3abc

Expand

(100 - a)(100 - b)(100 - c)

using

a + b + c = 300:

the condition says exactly that this product vanishes

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

\begin{gathered} (100 - a)(100 - b)(100 - c) \\ = 10^6 - 10^4 (a + b + c) \\ \quad {}+ 100q - p \\ = (100q - p) - 2 \cdot 10^6, \end{gathered}

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.

Problem 11 in other years