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

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==Solution 2 (drawing)==
 
==Solution 2 (drawing)==
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Draw a diagram (optionally with grid paper and/or a ruler), then simply count the number of tiles the path crosses, to get \boxed{\textbf{(C) }26}$ as in Solution 1.
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We draw a diagram (optionally with grid paper and/or a ruler), then simply count the number of tiles the path crosses. To make this slightly easier, we can divide the full grid into <math>4</math> sections, and just draw one of these <math>5</math> feet by <math>8.5</math> feet sections.
  
 
<asy>
 
<asy>
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</asy>
 
</asy>
  
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While it appears that the line we drew comes very close to several points, we know that since <math>10</math> and <math>17</math> are relatively prime, it will not actually pass through any of these points, so the total squares will be the same regardless of which side we count. If we sum the diagram, we get <math>13</math> squares, for a total of <math>26</math>.
+
While it appears that the line we drew comes very close to several points, we know that since <math>10</math> and <math>17</math> are relatively prime, it will not actually pass through any of these points, so the total number of squares crossed will be the same regardless of which side we count. If we count from the diagram, we get <math>13</math> squares, giving a total of <math>2 \cdot 13 = \boxed{\textbf{(C) }26}</math>.
  
 
==See Also==
 
==See Also==

Revision as of 23:38, 26 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 (drawing)

We draw a diagram (optionally with grid paper and/or a ruler), then simply count the number of tiles the path crosses. To make this slightly easier, we can divide the full grid into $4$ sections, and just draw one of these $5$ feet by $8.5$ feet sections.

[asy] unitsize(20); for(int i =0; i<= 7; ++i) { for(int j =0; j<= 4; ++j) { draw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--cycle); } for(int k =0; k<= 4; ++k) { draw((8,k)--(8.5,k)--(8.5,k+1)--(8,k+1)--cycle); } } draw((0,5)--(8.5,0)--cycle); [/asy]

While it appears that the line we drew comes very close to several points, we know that since $10$ and $17$ are relatively prime, it will not actually pass through any of these points, so the total number of squares crossed will be the same regardless of which side we count. If we count from the diagram, we get $13$ squares, giving a total of $2 \cdot 13 = \boxed{\textbf{(C) }26}$.

See Also

2019 AMC 10A (Problems • Answer Key • Resources)
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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