Difference between revisions of "2013 AMC 8 Problems/Problem 22"
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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. An effective way to verify this is to try a small case, i.e. a <math>2 \times 3</math> grid of toothpicks. Thus, our answer is <math>61\cdot 32 + 33 \cdot 60 = \boxed{\textbf{(E)}\ 3932}</math>. | 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. An effective way to verify this is to try a small case, i.e. a <math>2 \times 3</math> grid of toothpicks. Thus, our answer is <math>61\cdot 32 + 33 \cdot 60 = \boxed{\textbf{(E)}\ 3932}</math>. | ||
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{{AMC8 box|year=2013|num-b=21|num-a=23}} | {{AMC8 box|year=2013|num-b=21|num-a=23}} | ||
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Revision as of 20:49, 3 January 2024
Problem
Toothpicks are used to make a grid that is 
toothpicks long and 
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]](http://latex.artofproblemsolving.com/c/6/1/c610668e8c5623b4247005f01744cde6726b65a9.png)
Video Solution for Problems 21-25
Video Solution
https://youtu.be/nNDdkv_zfOo ~savannahsolver
Solution
There are 
vertical columns with a length of toothpicks, and there are 
horizontal rows with a length of toothpicks. An effective way to verify this is to try a small case, i.e. a 
grid of toothpicks. Thus, our answer is 
.
See Also :)
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
