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

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Using [[Simon's Favorite Factoring Trick]], we rewrite as <math>[(x+7)(y-3)]^n = (x+7)^n(y-3)^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 \ge 1996</math>, the smallest square after <math>1996</math> is <math>2025 = 45^2</math>, so our answer is <math>45 - 1 = \boxed{044}</math>.
 
Using [[Simon's Favorite Factoring Trick]], we rewrite as <math>[(x+7)(y-3)]^n = (x+7)^n(y-3)^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 \ge 1996</math>, the smallest square after <math>1996</math> is <math>2025 = 45^2</math>, so our answer is <math>45 - 1 = \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</math> --> <math>n = \boxed{044}</math>.
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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>.
  
 
== See also ==
 
== See also ==

Revision as of 16:12, 3 January 2018

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

See also

1996 AIME (ProblemsAnswer KeyResources)
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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