1999 AMC 12 Problem 30

Attempt Problem 30 of the 1999 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1999 AMC 12 solutions, or check the answer key.

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30.

The number of ordered pairs of integers

(m, n)

for which

mn \ge 0

and

m^3 + n^3 + 99mn = 33^3

is equal to

2

3

33

35

99

Answer: D

Concepts:factoring Diophantine Equation

Difficulty rating: 2460

Small hint:

Since

99 = 3 \cdot 33,

the equation is

m^3 + n^3 + (-33)^3

- 3mn(-33) = 0

Big hint:

Use

x^3 + y^3 + z^3

- 3xyz = (x + y + z)

\cdot (x^2 + y^2 + z^2 - xy - yz - zx)

Solution:

Writing

z = -33,

the equation becomes

m^3 + n^3 + z^3 - 3mnz = 0,

which factors as

\begin{aligned} &(m + n - 33) \\ &\quad {}\cdot \scriptsize\left(m^2 + n^2 + 33^2 - mn + 33m + 33n\right) \\ &= 0. \end{aligned}

The second factor equals

\scriptsize\tfrac12\big[(m - n)^2 + (m + 33)^2 + (n + 33)^2\big],

which is

0

only at

(m, n) = (-33, -33);

this satisfies

mn \ge 0.

Otherwise

m + n = 33.

With

mn \ge 0

both are nonnegative, giving

(0, 33), (1, 32), \ldots, (33, 0),

which is

34

pairs. Together there are

35

solutions.

Thus, the correct answer is D.