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 1 (symmetric sums)

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 $mn \ge 0$ .

Solution 2 (Binomial theorem / symmetric sums)

Since we see $^3$ and multiples of $3$ , let's try the bominomial theorem:

$(m+n)^3 = m^3 + n^3 + 3(m+n)mn$ .

This is almost exactly the main premise of the problem, so next try $m+n=33$ :

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

Since we require $mn \geq 0$ , we have solutions for all $0 \leq m \leq 33$ , which has $\mathbf{ 34 }$ solutions.

If $\max(m, n) \geq 33$ , the left side must be greater than the right side, so these are the only possible positive solutions.

Next, we consider negative solutions. (Also, the multiple choice options confirm that we are missing a solution. Further, since the remaining options are all odd, we get a hint that, due to symmetry, there must be an odd number of solutions with $m=n$ .).

Consider the case $m=n$ : $2m^3 + 99m^2 = 33^3$ , equivalent to $m^2(2m+99) = 33^3$ . If you don't immediately guess $m=-33$ (which works), observe that $m \equiv 0 \pmod {11}$ and also $\pmod 3$ . Arithmetic shows that $(-33,-33)$ is a solution for this case, and no other values of $m$ give $m^2(2m+99) \geq 0$ . So we have $\mathbf{1}$ additional solution.

This gives the correct answer of $34+1=\boxed{\textbf{(D) } 35}$ , but we haven't yet proved there are no other solutions..

Finally, we consider $m \neq n$ , when both are negative. This forces $m^3 \equiv -(n^3)$ in both $\pmod {11}$ and $\pmod 3$ , which implies (by computing all the modular cubes) that $m \equiv -n$ . This further implies that $m \equiv -n \pmod {33}$ , due to the modular (Chinese) Remainder Theorem.

But as $|m-n|$ increases $m^3+n^3$ grows more negative while $99mn$ shrinks toward 0. Thus the sum cannot remain constant, unless one of $m$ or $n$ jumps by 33.

There are a few remaining possibilities to check before $m^3$ grows far too large, but at this point we can make a nearly confident guess that $\textbf{(E) } 99$ is impossible. :-/

1999 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 29	Followed by Last problem
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

Page

Toolbox

Search

1999 AHSME Problems/Problem 30

Contents

Problem

Solution 1 (symmetric sums)

Solution 2 (Binomial theorem / symmetric sums)

See also