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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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 +
== Solution FASTER ==
 +
 +
the floor of the square root of 1996
 +
easily 44
  
 
== See also ==
 
== See also ==

Revision as of 17:27, 29 March 2020

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, $k = 45\rightarrow n = \boxed{044}$.

Solution FASTER

the floor of the square root of 1996 easily 44

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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