Difference between revisions of "2003 AMC 12A Problems/Problem 8"

Revision as of 18:01, 7 July 2020

The following problem is from both the 2003 AMC 12A #8 and 2003 AMC 10A #8, so both problems redirect to this page.

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

Solution 1

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 $\boxed{\mathrm{(E)}\ \frac{1}{2}}$ .

Solution 2

Testing all numbers less than $7$ , numbers $1, 2, 3, 4, 5$ , and $6$ divide $60$ . The prime factorization of $60$ is $2^2\cdot 3 \cdot 5$ . Using the formula for the number of divisors, the total number of divisors of $60$ is $(3)(2)(2) = 12$ . Therefore, our desired probability is $\frac{6}{12} = \boxed{\mathrm{(E)}\ \frac{1}{2}}$

Solution 3

This is not too bad with casework. Notice that $1*60=2*30=3*20=4*15=5*12=6*10=60$ . Hence, $60$ has $12$ factors, of which $6$ are less than $7$ . Thus, the answer is $\frac{6}{12} = \boxed{\mathrm{(E)}\ \frac{1}{2}}$ .

Solution by franzliszt

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

2003 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 7	Followed by Problem 9
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 AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 20: / Line 20: @@
 === Solution 2===
 Testing all numbers less than <math>7</math>, numbers <math>1, 2, 3, 4, 5</math>, and <math>6</math> divide <math>60</math>. The prime factorization of <math>60</math> is <math>2^2\cdot 3 \cdot 5</math>. Using the formula for the number of divisors, the total number of divisors of <math>60</math> is <math>(3)(2)(2) = 12</math>. Therefore, our desired probability is <math>\frac{6}{12} = \boxed{\mathrm{(E)}\ \frac{1}{2}}</math>
+===Solution 3===
+This is not too bad with casework. Notice that <math>1*60=2*30=3*20=4*15=5*12=6*10=60</math>. Hence, <math>60</math> has <math>12</math> factors, of which <math>6</math> are less than <math>7</math>. Thus, the answer is <math>\frac{6}{12} = \boxed{\mathrm{(E)}\ \frac{1}{2}}</math>.
+Solution by franzliszt
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