Difference between revisions of "1996 AIME Problems/Problem 3"
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== Solution == | == Solution == | ||
| − | Using | + | 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 20:54, 24 September 2007
Problem
Find the smallest positive integer 
for which the expansion of 
, after like terms have been collected, has at least 1996 terms.
Solution
Using Simon's Favorite Factoring Trick, rewrite as ![$[(x-3)(y-7)]^n = (x-3)^n(y-7)^n$](http://latex.artofproblemsolving.com/7/6/1/7615ccde29f986b3005f27f058b4d67226fb587b.png)
. Both binomial expansions will contain 
non-like terms; their product will contain 
terms, as each term will have an unique power of 
or 
and so none of the terms will need to be collected. Hence 
, the smallest square after 
is 
, so our answer is 
.
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 | ||
