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Difference between revisions of "2002 AIME II Problems/Problem 4"

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When <math>n>1</math>, the path of blocks has <math>6(n-1)</math> blocks total in it. When <math>n=1</math>, there is just one lonely block. Thus, the area of the garden enclosed by the path when <math>n=202</math> is
 
When <math>n>1</math>, the path of blocks has <math>6(n-1)</math> blocks total in it. When <math>n=1</math>, there is just one lonely block. Thus, the area of the garden enclosed by the path when <math>n=202</math> is
  
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<cmath>(1+6+12+18+\cdots +1200)A=(1+6(1+2+3...+200))A</cmath>
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<cmath>(1+6+12+18+\cdots +1200)A=(1+6(1+2+3...+200))A</cmath>,
  
−
Then, because n(n+1)/2 is equal to the sum of the first n integers,
+
where <math>A</math> is the area of one block. Then, because <math>n(n+1)/2</math> is equal to the sum of the first <math>n</math> integers:
  
−
<cmath>(1+6(1+2+3...+200))=(1+6((200)(201)/2))A=120601A</cmath>
+
<cmath>(1+6(1+2+3...+200))=(1+6((200)(201)/2))A=120601A</cmath>.
  
−
where <math>A</math> is the area of one block. Since <math>A=\dfrac{3\sqrt{3}}{2}</math>, the area of the garden is
+
Since <math>A=\dfrac{3\sqrt{3}}{2}</math>, the area of the garden is
  
 
<cmath>120601\cdot \dfrac{3\sqrt{3}}{2}=\dfrac{361803\sqrt{3}}{2}</cmath>.
 
<cmath>120601\cdot \dfrac{3\sqrt{3}}{2}=\dfrac{361803\sqrt{3}}{2}</cmath>.

Revision as of 16:51, 14 December 2015

Problem

Patio blocks that are hexagons $1$ unit on a side are used to outline a garden by placing the blocks edge to edge with $n$ on each side. The diagram indicates the path of blocks around the garden when $n=5$.

AIME 2002 II Problem 4.gif

If $n=202$, then the area of the garden enclosed by the path, not including the path itself, is $m\left(\sqrt3/2\right)$ square units, where $m$ is a positive integer. Find the remainder when $m$ is divided by $1000$.

Solution

When $n>1$, the path of blocks has $6(n-1)$ blocks total in it. When $n=1$, there is just one lonely block. Thus, the area of the garden enclosed by the path when $n=202$ is

\[(1+6+12+18+\cdots +1200)A=(1+6(1+2+3...+200))A\],

where $A$ is the area of one block. Then, because $n(n+1)/2$ is equal to the sum of the first $n$ integers:

\[(1+6(1+2+3...+200))=(1+6((200)(201)/2))A=120601A\].

Since $A=\dfrac{3\sqrt{3}}{2}$, the area of the garden is

\[120601\cdot \dfrac{3\sqrt{3}}{2}=\dfrac{361803\sqrt{3}}{2}\].

$m=361803$, $\dfrac{m}{1000}=361$ Remainder $\boxed{803}$.

See also

2002 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 3 		Followed by
Problem 5
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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