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

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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>.
 
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>.
  
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-superplayer24
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-pearlseal34
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==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}}

Latest revision as of 15:49, 6 September 2025

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 $\boxed{\textbf{(E)}\ 3932}$.

-pearlseal34

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

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