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

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(added problem and solution)
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== Problem ==
 
== Problem ==
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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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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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<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>
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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>
  
 
== Solution ==
 
== Solution ==
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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>.  
+
Let the squares be labeled <math>A</math>, <math>B</math>, <math>C</math>, and <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>.  
+
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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Clearly, there are no positive factors of <math>60</math> between <math>7</math> and <math>\sqrt{60}</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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Therefore half of the positive factors will be less than <math>7</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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So the answer is <math>\frac{1}{2} \Rightarrow E</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.
 +
 
 +
Thus the answer is <math>6 \Rightarrow E</math>.  
  
 
== See Also ==
 
== See Also ==
 
*[[2003 AMC 10A Problems]]
 
*[[2003 AMC 10A Problems]]
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*[[2003 AMC 10A Problems/Problem 7|Previous Problem]]
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*[[2003 AMC 10A Problems/Problem 9|Previous Problem]]
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*[[2003 AMC 10A Problems/Problem 9|Next Problem]]
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*[[2003 AMC 10A Problems/Problem 11|Next Problem]]
  
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[[Category:Introductory Number Theory Problems]]
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[[Category:Introductory Geometry Problems]]

Revision as of 19:02, 4 November 2006

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?

$\mathrm{(A) \ } 2\qquad \mathrm{(B) \ } 3\qquad \mathrm{(C) \ } 4\qquad \mathrm{(D) \ } 5\qquad \mathrm{(E) \ } 6$

Solution

Let the squares be labeled $A$, $B$, $C$, and $D$.

When the polygon is folded, the "right" edge of square $A$ becomes adjacent to the "bottom edge" of square $C$, and the "bottom" edge of square $A$ becomes adjacent to the "bottom" edge of square $D$.

So, any "new" square that is attached to those edges will prevent the polygon from becoming a cube with one face missing.

Therefore, squares $1$, $2$, and $3$ will prevent the polygon from becoming a cube with one face missing.

Squares $4$, $5$, $6$, $7$, $8$, and $9$ will allow the polygon to become a cube with one face missing when folded.

Thus the answer is $6 \Rightarrow E$.

See Also

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