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

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== Solution ==
 
== Solution ==
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Using [[Simon's Favorite Factoring Trick]], rewrite as <math>[(x-3)(y-7)]^n = (x-3)^n(y-7)^n</math>. Both [[binomial expansion]]s will contain <math>n+1</math> non-like terms; their product will contain <math>(n+1)^2</math> terms, as each term will have an unique power of <math>x</math> or <math>y</math> and so none of the terms will need to be collected. Hence <math>(n+1)^2 > 1996</math>, the smallest square after <math>1996</math> is <math>2025 = 45^2</math>, so our answer is <math>45 - 1 = 044</math>.
+
Using [[Simon's Favorite Factoring Trick]], rewrite as <math>[(x-3)(y+7)]^n = (x-3)^n(y+7)^n</math>. Both [[binomial expansion]]s will contain <math>n+1</math> non-like terms; their product will contain <math>(n+1)^2</math> terms, as each term will have an unique power of <math>x</math> or <math>y</math> and so none of the terms will need to be collected. Hence <math>(n+1)^2 > 1996</math>, the smallest square after <math>1996</math> is <math>2025 = 45^2</math>, so our answer is <math>45 - 1 = 044</math>.
  
 
== See also ==
 
== See also ==

Revision as of 22:13, 5 January 2008

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, rewrite as $[(x-3)(y+7)]^n = (x-3)^n(y+7)^n$. 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 $(n+1)^2 > 1996$, the smallest square after $1996$ is $2025 = 45^2$, so our answer is $45 - 1 = 044$.

See also

1996 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 2 		Followed by
Problem 4
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
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