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

Revision as of 20:59, 23 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 $26$ . Note: We can cut the floor into 4 pieces and draw a diagram for only one piece that is $5$ feet by $8.5$ feet.

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

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 squares will be the same regardless of which side we count. If we sum the diagram, we get $13$ squares, for a total of $26$ .

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

Page

Toolbox

Search

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

Revision as of 20:59, 23 February 2019

Contents

Problem

Solution 1

Solution 2 (Draw it out)

See Also

@@ Line 12: / Line 12: @@
 ==Solution 2 (Draw it out)==
 Draw it out using grid paper and a ruler.  Carefully counting the squares gives us <math>26</math>.
+Note: We can cut the floor into 4 pieces and draw a diagram for only one piece that is <math>5</math> feet by <math>8.5</math> feet.
+<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 <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>.
 ==See Also==