# Difference between revisions of "1996 AIME Problems/Problem 3"

## Problem

Find the smallest positive integer $n$ for which the expansion of $(xy-3x+7y-21)^n$, after like terms have been collected, has at least 1996 terms.

## Solution

Using Simon's Favorite Factoring Trick, we rewrite as $[(x+7)(y-3)]^n = (x+7)^n(y-3)^n$. Both binomial expansions will contain $n+1$ non-like terms; their product will contain $(n+1)^2$ terms, as each term will have an unique power of $x$ or $y$ and so none of the terms will need to be collected. Hence $(n+1)^2 \ge 1996$, the smallest square after $1996$ is $2025 = 45^2$, so our answer is $45 - 1 = \boxed{044}$.

Alternatively, when $n = k$, the exponents of $x$ or $y$ in $x^i y^i$ can be any integer between $0$ and $k$ inclusive. Thus, when $n=1$, there are $(2)(2)$ terms and, when $n = k$, there are $(k+1)^2$ terms. Therefore, we need to find the smallest perfect square that is greater than $1996$. From trial and error, we get $44^2 = 1936$ and $45^2 = 2025$. Thus, $k = 45\rightarrow n = \boxed{044}$.

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 