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1999 AMC 12 Problem 30

Below is the professionally curated solution for Problem 30 of the 1999 AMC 12, 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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Concepts:factoringDiophantine Equation

Difficulty rating: 2460

30.

The number of ordered pairs of integers (m,n)(m, n) for which mn≥0mn \ge 0 and

m3+n3+99mn=333m^3 + n^3 + 99mn = 33^3

is equal to

22

33

3333

3535

9999

Solution:

Writing z=−33,z = -33, the equation becomes m3+n3+z3−3mnz=0,m^3 + n^3 + z^3 - 3mnz = 0, which factors as (m+n−33)(m2+n2+332−mn+33m+33n)=0. (m + n - 33)\left(m^2 + n^2 + 33^2 - mn + 33m + 33n\right) = 0.

The second factor equals 12[(m−n)2+(m+33)2+(n+33)2],\tfrac12\big[(m - n)^2 + (m + 33)^2 + (n + 33)^2\big], which is 00 only at (m,n)=(−33,−33);(m, n) = (-33, -33); this satisfies mn≥0.mn \ge 0.

Otherwise m+n=33.m + n = 33. With mn≥0mn \ge 0 both are nonnegative, giving (0,33),(1,32),…,(33,0),(0, 33), (1, 32), \ldots, (33, 0), which is 3434 pairs. Together there are 3535 solutions.

Thus, the correct answer is D.

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