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 "2003 AMC 10A Problems/Problem 23"

(Switched "upwards" and "downwards" in solution 2. There are always more upwards-facing triangles than downwards-facing triangles in each row.)
Line 1: Line 1:
 
== Problem ==
 
== Problem ==
−
A large [[equilateral triangle]] is constructed by using toothpicks to create rows of small equilateral triangles. For example, in the figure we have <math>3</math> rows of small congruent equilateral triangles, with <math>5</math> small triangles in the base row.  How many toothpicks would be needed to construct a large equilateral triangle if the base row of the triangle consists of <math>2003</math> small equilateral triangles?  
+
A large [[equilateral triangle]] is constructed by using toothpicks to create rows of small equilateral triangles. For example, in the figure, we have <math>3</math> rows of small congruent equilateral triangles, with <math>5</math> small triangles in the base row.  How many toothpicks would be needed to construct a large equilateral triangle if the base row of the triangle consists of <math>2003</math> small equilateral triangles?  
  
 
[[Image:2003amc10a23.gif]]
 
[[Image:2003amc10a23.gif]]
Line 19: Line 19:
  
 
===Solution 2===
 
===Solution 2===
−
We see that the bottom row of <math>2003</math> small triangles is formed from <math>1002</math> upward-facing triangles and <math>1001</math> downward-facing triangles. Since each upward-facing triangle uses three distinct toothpicks, and since the total number of upward-facing triangles is <math>1002+1001+...+1=\frac{1003\cdot1002}{2}=502503</math>, we have that the total number of toothpicks is <math>3\cdot 502503=\boxed{\mathrm{(C)}\ 1,507,509}</math>
+
We see that the bottom row of <math>2003</math> small triangles are formed from <math>1002</math> upward-facing triangles and <math>1001</math> downward-facing triangles. Since each upward-facing triangle uses three distinct toothpicks, and since the total number of upward-facing triangles is <math>1002+1001+...+1=\frac{1003\cdot1002}{2}=502503</math>, we have that the total number of toothpicks is <math>3\cdot 502503=\boxed{\mathrm{(C)}\ 1,507,509}</math>
  
 
===Solution 3===
 
===Solution 3===
−
Experimenting a bit we find that the number of toothpicks needs for a triangle with <math>1</math>, <math>2</math> and <math>3</math> rows is <math>1\cdot{3}</math>, <math>3\cdot{3}</math> and <math>6\cdot{3}</math> respectively. Since <math>1</math>, <math>3</math> and <math>6</math> are triangular numbers we know that depending on how many rows there are in the triangle, the number we multiply by <math>3</math> to find total no.toothpicks is the corresponding triangular number. Since the triangle in question has <math>2n-1=2003\implies{n=1002}</math> rows, we can use <math>\frac{n(n+1)}{2}</math> to find the triangular number for that row and multiply by <math>3</math>, hence finding the total no.toothpicks; this is just <math>\frac{3\cdot{1002}\cdot{1003}}{2}=3\cdot{501}\cdot{1003}=\boxed{\mathrm{(C)}\ 1,507,509}</math>.
+
Experimenting a bit we find that the number of toothpicks needs a triangle with <math>1</math>, <math>2</math> and <math>3</math> rows is <math>1\cdot{3}</math>, <math>3\cdot{3}</math> and <math>6\cdot{3}</math> respectively. Since <math>1</math>, <math>3</math> and <math>6</math> are triangular numbers we know that depending on how many rows there are in the triangle, the number we multiply by <math>3</math> to find total no.toothpicks is the corresponding triangular number. Since the triangle in question has <math>2n-1=2003\implies{n=1002}</math> rows, we can use <math>\frac{n(n+1)}{2}</math> to find the triangular number for that row and multiply by <math>3</math>, hence finding the total no.toothpicks; this is just <math>\frac{3\cdot{1002}\cdot{1003}}{2}=3\cdot{501}\cdot{1003}=\boxed{\mathrm{(C)}\ 1,507,509}</math>.
 +
===Note===
 +
In the final step of the problem, we know that the units digit of the answer must <math>9</math>, so the only answer choice applicable must be <math>\boxed{\mathrm{(C)}\ 1,507,509}</math>. This saves you the time it takes to compute <math>3\cdot{501}\cdot{1003}</math>.
  
 
== See Also ==
 
== See Also ==

Revision as of 22:54, 29 December 2017

Problem

A large equilateral triangle is constructed by using toothpicks to create rows of small equilateral triangles. For example, in the figure, we have $3$ rows of small congruent equilateral triangles, with $5$ small triangles in the base row. How many toothpicks would be needed to construct a large equilateral triangle if the base row of the triangle consists of $2003$ small equilateral triangles?

2003amc10a23.gif

$\mathrm{(A) \ } 1,004,004 \qquad \mathrm{(B) \ } 1,005,006 \qquad \mathrm{(C) \ } 1,507,509 \qquad \mathrm{(D) \ } 3,015,018 \qquad \mathrm{(E) \ } 6,021,018$

Solution

Solution 1

There are $1+3+5+...+2003=1002^{2}=1004004$ small equilateral triangles.

Each small equilateral triangle needs $3$ toothpicks to make it.

But, each toothpick that isn't one of the $1002\cdot3=3006$ toothpicks on the outside of the large equilateral triangle is a side for $2$ small equilateral triangles.

So, the number of toothpicks on the inside of the large equilateral triangle is $\frac{10040004\cdot3-3006}{2}=1504503$

Therefore the total number of toothpicks is $1504503+3006=\boxed{\mathrm{(C)}\ 1,507,509}$

Solution 2

We see that the bottom row of $2003$ small triangles are formed from $1002$ upward-facing triangles and $1001$ downward-facing triangles. Since each upward-facing triangle uses three distinct toothpicks, and since the total number of upward-facing triangles is $1002+1001+...+1=\frac{1003\cdot1002}{2}=502503$, we have that the total number of toothpicks is $3\cdot 502503=\boxed{\mathrm{(C)}\ 1,507,509}$

Solution 3

Experimenting a bit we find that the number of toothpicks needs a triangle with $1$, $2$ and $3$ rows is $1\cdot{3}$, $3\cdot{3}$ and $6\cdot{3}$ respectively. Since $1$, $3$ and $6$ are triangular numbers we know that depending on how many rows there are in the triangle, the number we multiply by $3$ to find total no.toothpicks is the corresponding triangular number. Since the triangle in question has $2n-1=2003\implies{n=1002}$ rows, we can use $\frac{n(n+1)}{2}$ to find the triangular number for that row and multiply by $3$, hence finding the total no.toothpicks; this is just $\frac{3\cdot{1002}\cdot{1003}}{2}=3\cdot{501}\cdot{1003}=\boxed{\mathrm{(C)}\ 1,507,509}$.

Note

In the final step of the problem, we know that the units digit of the answer must $9$, so the only answer choice applicable must be $\boxed{\mathrm{(C)}\ 1,507,509}$. This saves you the time it takes to compute $3\cdot{501}\cdot{1003}$.

See Also

2003 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 22 		Followed by
Problem 24
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 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=2003_AMC_10A_Problems/Problem_23&oldid=89259"