Difference between revisions of "2013 AMC 8 Problems/Problem 22"
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==Solution== | ==Solution== | ||
− | 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)}\ | + | 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>. |
==See Also== | ==See Also== | ||
{{AMC8 box|year=2013|num-b=21|num-a=23}} | {{AMC8 box|year=2013|num-b=21|num-a=23}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 20:32, 27 November 2013
Problem
Toothpicks are used to make a grid that is 60 toothpicks long and 32 toothpicks wide. How many toothpicks are used altogether?
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 |
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