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 "2003 AMC 10A Problems/Problem 8"

(→‎See Also: fixed link)
(fixed typo)
Line 1: Line 1:
 
== Problem ==
 
== Problem ==
−
The polygon enclosed by the solid lines in the figure consists of 4 congruent squares joined edge-to-edge. One more congruent square is attached to an edge at one of the nine positions indicated. How many of the nine resulting polygons can be folded to form a cube with one face missing?
+
What is the probability that a randomly drawn positive factor of <math>60</math> is less than <math>7</math>
  
−
<math> \mathrm{(A) \ } 2\qquad \mathrm{(B) \ } 3\qquad \mathrm{(C) \ } 4\qquad \mathrm{(D) \ } 5\qquad \mathrm{(E) \ } 6 </math>
+
<math> \mathrm{(A) \ } \frac{1}{10}\qquad \mathrm{(B) \ } \frac{1}{6}\qquad \mathrm{(C) \ } \frac{1}{4}\qquad \mathrm{(D) \ } \frac{1}{3}\qquad \mathrm{(E) \ } \frac{1}{2} </math>
  
 
== Solution ==
 
== Solution ==
−
Let the squares be labeled <math>A</math>, <math>B</math>, <math>C</math>, and <math>D</math>.
+
For a positive number <math>n</math> which is not a perfect square, exactly half of the positive factors will be less than <math>\sqrt{n}</math>.  
  
−
When the polygon is folded, the "right" edge of square <math>A</math> becomes adjacent to the "bottom edge" of square <math>C</math>, and the "bottom" edge of square <math>A</math> becomes adjacent to the "bottom" edge of square <math>D</math>.  
+
Since <math>60</math> is not a perfect square, half of the positive factors of <math>60</math> will be less than <math>\sqrt{60}\approx 7.746</math>.  
  
−
So, any "new" square that is attached to those edges will prevent the polygon from becoming a cube with one face missing.  
+
Clearly, there are no positive factors of <math>60</math> between <math>7</math> and <math>\sqrt{60}</math>.  
  
−
Therefore, squares <math>1</math>, <math>2</math>, and <math>3</math> will prevent the polygon from becoming a cube with one face missing.
+
Therefore half of the positive factors will be less than <math>7</math>.  
  
−
Squares  <math>4</math>, <math>5</math>, <math>6</math>, <math>7</math>, <math>8</math>, and <math>9</math> will allow the polygon to become a cube with one face missing when folded.
+
So the answer is <math>\frac{1}{2} \Rightarrow E</math>.  
−
 
 
−
Thus the answer is <math>6 \Rightarrow E</math>.  
 
  
 
== See Also ==
 
== See Also ==
Line 22: Line 20:
 
*[[2003 AMC 10A Problems/Problem 9|Next Problem]]
 
*[[2003 AMC 10A Problems/Problem 9|Next Problem]]
  
−
[[Category:Introductory Geometry Problems]]
+
[[Category:Introductory Number Theory Problems]]

Revision as of 19:08, 4 November 2006

Problem

What is the probability that a randomly drawn positive factor of $60$ is less than $7$

$\mathrm{(A) \ } \frac{1}{10}\qquad \mathrm{(B) \ } \frac{1}{6}\qquad \mathrm{(C) \ } \frac{1}{4}\qquad \mathrm{(D) \ } \frac{1}{3}\qquad \mathrm{(E) \ } \frac{1}{2}$

Solution

For a positive number $n$ which is not a perfect square, exactly half of the positive factors will be less than $\sqrt{n}$.

Since $60$ is not a perfect square, half of the positive factors of $60$ will be less than $\sqrt{60}\approx 7.746$.

Clearly, there are no positive factors of $60$ between $7$ and $\sqrt{60}$.

Therefore half of the positive factors will be less than $7$.

So the answer is $\frac{1}{2} \Rightarrow E$.

See Also

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2003_AMC_10A_Problems/Problem_8&oldid=10885"