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1999 AHSME Problems/Problem 30

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 $m = n = - 33$ (by the Trivial Inequality). Thus, there are $35 \Longrightarrow \mathrm{(D)}$ solutions satisfying $mn \ge 0$.

See also

1999 AHSME (ProblemsAnswer KeyResources)
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