Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2019 AMC 10A Problems/Problem 10"
m (→‎Solution 1)
(4 intermediate revisions by 2 users not shown)
Line 9: Line 9:
  
 
''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.
 
''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.
−
 
  
 
One can see why it is gcd(a,b) due to slope
 
One can see why it is gcd(a,b) due to slope
 
~Williamgolly
 
~Williamgolly
 +
 +
''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.
 +
~aliciawu
  
 
==Solution 2 (drawing)==
 
==Solution 2 (drawing)==
Line 44: Line 46:
  
  
−
== Video Solution ==
+
== Video Solution by Omega Learn ==
 
https://youtu.be/zfChnbMGLVQ?t=843
 
https://youtu.be/zfChnbMGLVQ?t=843
  

Revision as of 22:31, 23 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 $2 \cdot 13 = \boxed{\textbf{(C) }26}$.

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10A_Problems/Problem_10&oldid=225624"