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

Revision as of 10:15, 15 January 2008

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 $\frac{1}{2} \Rightarrow E$ .

2003 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 7	Followed by Problem 9
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All AMC 10 Problems and Solutions

@@ Line 16: / Line 16: @@
 == See Also ==
-*[[2003 AMC 10A Problems]]
+{{AMC10 box|year=2003|ab=A|num-b=7|num-a=9}}
-*[[2003 AMC 10A Problems/Problem 7|Previous Problem]]
-*[[2003 AMC 10A Problems/Problem 9|Next Problem]]
 [[Category:Introductory Number Theory Problems]]

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

Revision as of 10:15, 15 January 2008

Problem

Solution

See Also