Difference between revisions of "1996 AIME Problems/Problem 3"
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Alternatively, when <math>n = k</math>, the exponents of <math>x</math> or <math>y</math> in <math>x^i y^i</math> can be any integer between <math>0</math> and <math>k</math> inclusive. Thus, when <math>n=1</math>, there are <math>(2)(2)</math> terms and, when <math>n = k</math>, there are <math>(k+1)^2</math> terms. Therefore, we need to find the smallest perfect square that is greater than <math>1996</math>. From trial and error, we get <math>44^2 = 1936</math> and <math>45^2 = 2025</math>. Thus, <math>k = 45\rightarrow n = \boxed{044}</math>. | Alternatively, when <math>n = k</math>, the exponents of <math>x</math> or <math>y</math> in <math>x^i y^i</math> can be any integer between <math>0</math> and <math>k</math> inclusive. Thus, when <math>n=1</math>, there are <math>(2)(2)</math> terms and, when <math>n = k</math>, there are <math>(k+1)^2</math> terms. Therefore, we need to find the smallest perfect square that is greater than <math>1996</math>. From trial and error, we get <math>44^2 = 1936</math> and <math>45^2 = 2025</math>. Thus, <math>k = 45\rightarrow n = \boxed{044}</math>. | ||
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== See also == | == See also == |
Latest revision as of 10:31, 19 April 2020
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, we rewrite as . 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 .
Alternatively, when , the exponents of or in can be any integer between and inclusive. Thus, when , there are terms and, when , there are terms. Therefore, we need to find the smallest perfect square that is greater than . From trial and error, we get and . Thus, .
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.