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

Latest revision as of 14:50, 27 August 2024

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.

One can see why it is gcd(a,b) due to slope ~Williamgolly

Comment: The above note defines a, b incorrectly. One counter example is a 17x9 grid, which should result in 25 tiles. However, $\text{gcd}(16, 8) = 8$ . Here $a+b-\gcd(a,b)$ is correct when a = 17 and b = 10. ~aliciawu

Solution 2 (drawing)

We can also draw a diagram or scale model of the entire rectangular floor (optionally with grid paper and/or a ruler so it will be to scale), 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.

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

Though it may appear that the line we drew comes very close to several points, we know that since $10$ and $17$ are relatively prime (numbers where the only positive integer that divides both of them is 1, a.k.a. numbers with a gcd of 1), the line 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 the number of squares the line passes through using the diagram, we get $13$ squares. We can then multiply this by 2 to find out the total number of squares the bug passes through on the rectangular floor giving us a total of $2 \cdot 13 = \boxed{\textbf{(C) }26}$ .

Solution 3 (Under 1 Minute)

We can see that the big $10$ by $17$ rectangle can be split into $2$ smaller $5$ by $17$ rectangles. This means that the number of small rectangles is divisible by $2$ . This also means that the number of small rectangles in the $5$ by $17$ rectangles is an odd number since it can't be divided into any smaller triangles. Thus the only number that meets the requirements is $\boxed{\textbf{(C) } 26}$ .

~Continuous_Pi

Video Solution

https://www.youtube.com/watch?v=qN1g7Vt5SCg

https://youtu.be/Z-sUMqZH0j4

~savannahsolver

Video Solution by Omega Learn

https://youtu.be/zfChnbMGLVQ?t=843

~ pi_is_3.14

@@ Line 5: / Line 5: @@
 <math>\textbf{(A) } 17 \qquad\textbf{(B) } 25 \qquad\textbf{(C) } 26 \qquad\textbf{(D) } 27 \qquad\textbf{(E) } 28</math>
-==Solution==
+==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.
-Note: The general formula for this is <math>a+b-\gcd(a,b)</math>.
+''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.
+One can see why it is gcd(a,b) due to slope
+~Williamgolly
-==Solution 2 (Draw it out)==
+''Comment'': The above note defines a, b incorrectly. One counter example is a 17x9 grid, which should result in 25 tiles. However, <math>\text{gcd}(16, 8) = 8</math>. Here <math>a+b-\gcd(a,b)</math> is correct when a = 17 and b = 10.
-Draw it out using grid paper and a ruler.  Carefully counting the squares gives us 26.
+~aliciawu
+==Solution 2 (drawing)==
+We can also draw a diagram or scale model of the entire rectangular floor (optionally with grid paper and/or a ruler so it will be to scale), 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>
+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>
+Though it may appear 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 (numbers where the only positive integer that divides both of them is 1, a.k.a. numbers with a gcd of 1), the line 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 the number of squares the line passes through using the diagram, we get <math>13</math> squares. We can then multiply this by 2 to find out the total number of squares the bug passes through on the rectangular floor giving us a total of <math>2 \cdot 13 = \boxed{\textbf{(C) }26}</math>.
+==Solution 3 (Under 1 Minute)==
+We can see that the big <math>10</math> by <math>17</math> rectangle can be split into <math>2</math> smaller <math>5</math> by <math>17</math> rectangles. This means that the number of small rectangles is divisible by <math>2</math>. This also means that the number of small rectangles in the <math>5</math> by <math>17</math> rectangles is an odd number since it can't be divided into any smaller triangles. Thus the only number that meets the requirements is <math>\boxed{\textbf{(C) } 26}</math>.
+~Continuous_Pi
+==Video Solution==
+https://www.youtube.com/watch?v=qN1g7Vt5SCg
+https://youtu.be/Z-sUMqZH0j4
+~savannahsolver
+== Video Solution by Omega Learn ==
+https://youtu.be/zfChnbMGLVQ?t=843
+~ pi_is_3.14
 ==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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