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1999 AHSME Problems/Problem 30

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Problem

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

\mathrm{(A) \ }2 \qquad \mathrm{(B) \ } 3\qquad \mathrm{(C) \ } 33\qquad \mathrm{(D) \ }35 \qquad \mathrm{(E) \ } 99

Solution

We recall the factorization (see elementary symmetric sums)

x^3 + y^3 + z^3 - 3xyz = \frac12\cdot(x + y + z)\cdot((x - y)^2 + (y - z)^2 + (z - x)^2)

Setting x = m,y = n,z = - 33, we have that either m + n - 33 = 0 or m = m = - 33 (by the Trivial Inequality). Thus, there are 35 \Longrightarrow \mathrm{(D)} solutions satisfying mn \ge 0.

See also

1999 AHSME (Problems)
Preceded by
Problem 29
Followed by
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