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 "2007 AMC 8 Problems/Problem 11"

(→‎Video Solution by WhyMath)
(→‎Solution)
Line 38: Line 38:
 
== Solution ==
 
== Solution ==
  
−
We first notice that tile III has a <math>0</math> on the bottom and a <math>5</math> on the right side. Since no other tile has a <math>0</math> or a <math>5</math>, Tile III must be in rectangle <math>D</math>. Tile III also has a <math>1</math> on the left, so Tile IV must be in Rectangle <math>C</math>.
+
We first notice that tile III has a 0 on the bottom and a 5 on the right side. Since no other tile has a 0 or a 5 Tile III must be in rectangle D Tile III also has a 1 on the left, so Tile IV must be in Rectangle C
  
−
The answer is <math>\boxed{\textbf{(D)}}</math>
+
The answer is D
−
 
 
−
<center>[[Image:AMC8_2007_11S.png]]</center>
 
  
 
==Video Solution by Why Math==
 
==Video Solution by Why Math==

Revision as of 18:09, 1 August 2021

Problem

Tiles $I, II, III$ and $IV$ are translated so one tile coincides with each of the rectangles $A, B, C$ and $D$. In the final arrangement, the two numbers on any side common to two adjacent tiles must be the same. Which of the tiles is translated to Rectangle $C$?

[asy] size(400); defaultpen(linewidth(0.8)); path p=origin--(8,0)--(8,6)--(0,6)--cycle; draw(p^^shift(8.5,0)*p^^shift(8.5,10)*p^^shift(0,10)*p); draw(shift(20,2)*p^^shift(28,2)*p^^shift(20,8)*p^^shift(28,8)*p); label("8", (4,6+10), S); label("6", (4+8.5,6+10), S); label("7", (4,6), S); label("2", (4+8.5,6), S); label("I", (4,6+10), N); label("II", (4+8.5,6+10), N); label("III", (4,6), N); label("IV", (4+8.5,6), N); label("3", (0,3+10), E); label("4", (0+8.5,3+10), E); label("1", (0,3), E); label("9", (0+8.5,3), E); label("7", (4,10), N); label("2", (4+8.5,10), N); label("0", (4,0), N); label("6", (4+8.5,0), N); label("9", (8,3+10), W); label("3", (8+8.5,3+10), W); label("5", (8,3), W); label("1", (8+8.5,3), W); label("A", (24,10), N); label("B", (32,10), N); label("C", (24,4), N); label("D", (32,4), N);[/asy]

$\mathrm{(A)}\ I \qquad \mathrm{(B)}\ II \qquad \mathrm{(C)}\ III \qquad \mathrm{(D)}\ IV \qquad \mathrm{(E)}$ cannot be determined

Solution

We first notice that tile III has a 0 on the bottom and a 5 on the right side. Since no other tile has a 0 or a 5 Tile III must be in rectangle D Tile III also has a 1 on the left, so Tile IV must be in Rectangle C

The answer is D

Video Solution by Why Math

https://youtu.be/WobCYII7TRg

~savannahsolver

See Also

2007 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 10 		Followed by
Problem 12
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 AJHSME/AMC 8 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=2007_AMC_8_Problems/Problem_11&oldid=159388"