Difference between revisions of "2019 AMC 10A Problems/Problem 10"
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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 (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 | + | 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 (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 in the rectangle giving us a total of <math>2 \cdot 13 = \boxed{\textbf{(C) }26}</math>. |
==See Also== | ==See Also== | ||
Revision as of 17:45, 29 December 2019
Problem
A rectangular floor that is 
feet wide and 
feet long is tiled with 
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?
Solution 1
The number of tiles the bug visits is equal to 
plus the number of times it crosses a horizontal or vertical line. As it must cross 
horizontal lines and 
vertical lines, it must be that the bug visits a total of 
squares.
Note: The general formula for this is 
, 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 
(since 
), which is 
, but then you add 
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 
sections, and just draw one of these 
feet by 
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]](http://latex.artofproblemsolving.com/1/4/d/14d398aed55067e3db65683ddf5a5c7cb08d2fd6.png)
While it appears that the line we drew comes very close to several points, we know that since and 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 
squares. We can then multiply this by 2 to find out the total number of squares the bug passes through in the rectangle giving us a total of 
.
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
