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. 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. 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>. | \times 4</math>. |
Revision as of 17:48, 27 September 2024
Problem
Toothpicks are used to make a grid that is toothpicks long and toothpicks wide. How many toothpicks are used altogether?
Video Solution
https://youtu.be/nNDdkv_zfOo ~savannahsolver
Solution 1
There are vertical columns with a length of toothpicks, and there are horizontal rows with a length of toothpicks. You can verify this by trying a smaller case, i.e. a grid of toothpicks, with and .
Thus, our answer is .
~Note by Theraccoon: The person who posted this answer did not include their name.
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 which would be roughly .
-superplayer24
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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