Difference between revisions of "2013 AMC 8 Problems/Problem 22"

Latest revision as of 10:28, 16 November 2024

Problem

Toothpicks are used to make a grid that is $60$ toothpicks long and $32$ toothpicks wide. How many toothpicks are used altogether?

$[asy] picture corner; draw(corner,(5,0)--(35,0)); draw(corner,(0,-5)--(0,-35)); for (int i=0; i<3; ++i){for (int j=0; j>-2; --j){if ((i-j)<3){add(corner,(50i,50j));}}} draw((5,-100)--(45,-100)); draw((155,0)--(185,0),dotted+linewidth(2)); draw((105,-50)--(135,-50),dotted+linewidth(2)); draw((100,-55)--(100,-85),dotted+linewidth(2)); draw((55,-100)--(85,-100),dotted+linewidth(2)); draw((50,-105)--(50,-135),dotted+linewidth(2)); draw((0,-105)--(0,-135),dotted+linewidth(2));[/asy]$

$\textbf{(A)}\ 1920 \qquad \textbf{(B)}\ 1952 \qquad \textbf{(C)}\ 1980 \qquad \textbf{(D)}\ 2013 \qquad \textbf{(E)}\ 3932$

Video Solution

https://youtu.be/nNDdkv_zfOo ~savannahsolver

Solution 1

There are $61$ vertical columns with a length of $32$ toothpicks, and there are $33$ horizontal rows with a length of $60$ toothpicks, because $32$ and $60$ are the number of intervals. You can verify this by trying a smaller case, i.e. a $3 \times 4$ grid of toothpicks, with $3 \times 3$ and $2 \times 4$ .

Thus, our answer is $61\cdot 32 + 33 \cdot 60 = \boxed{\textbf{(E)}\ 3932}$ .

~Note by Theraccoon: The person who posted this answer did not include their name. Minor edit by ~NXC

Solution 2 - Common sense

With a quick mental calculation, 60 * 30 yields 1800, which is roughly where 4 of our 5 answer choices lie in. However, we can tell that each square would require at least 2 toothpicks that uniquely belong to itself, so the answer would be $60\cdot 30 \cdot 2$ which would be roughly $\boxed{\textbf{(E)}\ 3932}$ .

-superplayer24

@@ Line 16: / Line 16: @@
 <math>\textbf{(A)}\ 1920 \qquad \textbf{(B)}\ 1952 \qquad \textbf{(C)}\ 1980 \qquad \textbf{(D)}\ 2013 \qquad \textbf{(E)}\ 3932</math>
-==Video Solution for Problems 21-25==
-https://youtu.be/-mi3qziCuec
 ==Video Solution==
 https://youtu.be/nNDdkv_zfOo ~savannahsolver
-==Solution==
+==Solution 1==
-There are <math>61</math> vertical columns with a length of <math>32</math> toothpicks, and there are <math>33</math> horizontal rows with a length of <math>60</math> toothpicks. You can verify this by trying a smaller case, i.e. a <math>3 \times 4</math> grid of toothpicks, with <math>3 \times 3</math> and <math>2
+There are <math>61</math> vertical columns with a length of <math>32</math> toothpicks, and there are <math>33</math> horizontal rows with a length of <math>60</math> toothpicks, because <math>32</math> and <math>60</math> are the number of intervals. You can verify this by trying a smaller case, i.e. a <math>3 \times 4</math> grid of toothpicks, with <math>3 \times 3</math> and <math>2
   \times 4</math>.
 Thus, our answer is <math>61\cdot 32 + 33 \cdot 60 = \boxed{\textbf{(E)}\ 3932}</math>.
+~Note by Theraccoon: The person who posted this answer did not include their name. Minor edit by ~NXC
 ==Solution 2 - Common sense==
-With a quick mental calculation, <math>60 /cdot 30 </math>yields 1800, which is roughly where 4 of our 5 answer choices lie in. However, we can tell that each square would require at least 2 toothpicks that uniquely belong to itself, so the answer would be <math>60\cdot 30 \cdot 2</math> which would be roughly <math>\boxed{\textbf{(E)}\ 3932}</math>.
+With a quick mental calculation, 60 * 30 yields 1800, which is roughly where 4 of our 5 answer choices lie in. However, we can tell that each square would require at least 2 toothpicks that uniquely belong to itself, so the answer would be <math>60\cdot 30 \cdot 2</math> which would be roughly <math>\boxed{\textbf{(E)}\ 3932}</math>.
 -superplayer24

2013 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 21	Followed by Problem 23
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 AJHSME/AMC 8 Problems and Solutions

Page

Toolbox

Search