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Difference between revisions of "2003 AMC 10A Problems/Problem 8"

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== Problem ==
 
== Problem ==
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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?
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What is the probability that a randomly drawn positive factor of <math>60</math> is less than <math>7</math>
  
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<math> \mathrm{(A) \ } 2\qquad \mathrm{(B) \ } 3\qquad \mathrm{(C) \ } 4\qquad \mathrm{(D) \ } 5\qquad \mathrm{(E) \ } 6 </math>
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<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 ==
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Let the squares be labeled <math>A</math>, <math>B</math>, <math>C</math>, and <math>D</math>.
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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>.  
  
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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>.  
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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>.  
  
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So, any "new" square that is attached to those edges will prevent the polygon from becoming a cube with one face missing.  
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Clearly, there are no positive factors of <math>60</math> between <math>7</math> and <math>\sqrt{60}</math>.  
  
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Therefore, squares <math>1</math>, <math>2</math>, and <math>3</math> will prevent the polygon from becoming a cube with one face missing.
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Therefore half of the positive factors will be less than <math>7</math>.  
  
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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.
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So the answer is <math>\frac{1}{2} \Rightarrow E</math>.  
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Thus the answer is <math>6 \Rightarrow E</math>.  
 
  
 
== See Also ==
 
== See Also ==
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*[[2003 AMC 10A Problems/Problem 9|Next Problem]]
 
*[[2003 AMC 10A Problems/Problem 9|Next Problem]]
  
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[[Category:Introductory Geometry Problems]]
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[[Category:Introductory Number Theory Problems]]

Revision as of 20:08, 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

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