Difference between revisions of "2019 AMC 10A Problems/Problem 10"

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<math>\textbf{(A) } 17 \qquad\textbf{(B) } 25 \qquad\textbf{(C) } 26 \qquad\textbf{(D) } 27 \qquad\textbf{(E) } 28</math>
 
<math>\textbf{(A) } 17 \qquad\textbf{(B) } 25 \qquad\textbf{(C) } 26 \qquad\textbf{(D) } 27 \qquad\textbf{(E) } 28</math>
  
==Solution==
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==Solution 1==
 
The number of tiles the bug visits is equal to <math>1</math> plus the number of times it crosses a horizontal or vertical line.  As it must cross <math>16</math> horizontal lines and <math>9</math> vertical lines, it must be that the bug visits a total of <math>16+9+1 = \boxed{\textbf{(C) }26}</math> squares.
 
The number of tiles the bug visits is equal to <math>1</math> plus the number of times it crosses a horizontal or vertical line.  As it must cross <math>16</math> horizontal lines and <math>9</math> vertical lines, it must be that the bug visits a total of <math>16+9+1 = \boxed{\textbf{(C) }26}</math> squares.
  
Note: The general formula for this is <math>a+b-\gcd(a,b)</math>.
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''Note'': The general formula for this is <math>a+b-\gcd(a,b)</math>, because it is the number of vertical/horizontal lines crossed minus the number of corners crossed (to avoid double counting). In this particular problem, it was <math>16 + 9 - 1</math> (since <math>\text{gcd}(16,9) = 1</math>), which is <math>24</math>, but then you add <math>2</math> because the first tile and the last tile are counted, which in the general formula are not counted.
 
 
Edit: The general formula is that because it is the number of vert/horz lines crossed minus the number of corners crossed (b/c that would be double counting). In this particular problem, it was 16 + 9 - 1 (gcd of 16 and 9 is 1), which is 24, but then you add two because the first tile and the last tile are counted, which in the general formula are not counted.
 
  
 
==Solution 2 (Draw it out)==
 
==Solution 2 (Draw it out)==
Draw it out using grid paper and a ruler.  Carefully counting the squares gives us 26.
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Draw it out using grid paper and a ruler.  Carefully counting the squares gives us <math>26</math>.
  
 
==See Also==
 
==See Also==

Revision as of 21:11, 17 February 2019

Problem

A rectangular floor that is $10$ feet wide and $17$ feet long is tiled with $170$ one-foot square tiles. A bug walks from one corner to the opposite corner in a straight line. Including the first and the last tile, how many tiles does the bug visit?

$\textbf{(A) } 17 \qquad\textbf{(B) } 25 \qquad\textbf{(C) } 26 \qquad\textbf{(D) } 27 \qquad\textbf{(E) } 28$

Solution 1

The number of tiles the bug visits is equal to $1$ plus the number of times it crosses a horizontal or vertical line. As it must cross $16$ horizontal lines and $9$ vertical lines, it must be that the bug visits a total of $16+9+1 = \boxed{\textbf{(C) }26}$ squares.

Note: The general formula for this is $a+b-\gcd(a,b)$, because it is the number of vertical/horizontal lines crossed minus the number of corners crossed (to avoid double counting). In this particular problem, it was $16 + 9 - 1$ (since $\text{gcd}(16,9) = 1$), which is $24$, but then you add $2$ because the first tile and the last tile are counted, which in the general formula are not counted.

Solution 2 (Draw it out)

Draw it out using grid paper and a ruler. Carefully counting the squares gives us $26$.

See Also

2019 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 9
Followed by
Problem 11
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 AMC 10 Problems and Solutions

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