Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution)
(→‎Solution 2)
 
(3 intermediate revisions by 3 users not shown)
Line 6: Line 6:
 
If <math>n=202</math>, then the area of the garden enclosed by the path, not including the path itself, is <math>m\left(\sqrt3/2\right)</math> square units, where <math>m</math> is a positive integer. Find the remainder when <math>m</math> is divided by <math>1000</math>.
 
If <math>n=202</math>, then the area of the garden enclosed by the path, not including the path itself, is <math>m\left(\sqrt3/2\right)</math> square units, where <math>m</math> is a positive integer. Find the remainder when <math>m</math> is divided by <math>1000</math>.
  
−
== Solution ==
+
== Solution 1==
 
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
  
−
<cmath>(1+6+12+18+\cdots +1200)A=(1+6(1+2+3...+200))A</cmath>
+
<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>.
  
 
<math>m=361803</math>, <math>\dfrac{m}{1000}=361</math> Remainder <math>\boxed{803}</math>.
 
<math>m=361803</math>, <math>\dfrac{m}{1000}=361</math> Remainder <math>\boxed{803}</math>.
 +
 +
== Solution 2==
 +
Note that this is just the definition for a centered hexagonal number, and  the formula for <math>(n-1)^{th}</math> term is <math>3n(n+1)+1</math>. Applying this for <math>200</math> as we want the inner area gives <math>120601</math>. Then continue as above.
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=2002|n=II|num-b=3|num-a=5}}
 
{{AIME box|year=2002|n=II|num-b=3|num-a=5}}
 +
 +
[[Category: Intermediate Geometry Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 06:28, 13 September 2020

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 1

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 $\boxed{803}$.

Solution 2

Note that this is just the definition for a centered hexagonal number, and the formula for $(n-1)^{th}$ term is $3n(n+1)+1$. Applying this for $200$ as we want the inner area gives $120601$. Then continue as above.

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2002_AIME_II_Problems/Problem_4&oldid=133575"