2003 AMC 12A Problems/Problem 8

Revision as of 18:01, 7 July 2020 by Franzliszt (talk | contribs) (Solution)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
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

See Also

2003 AMC 10A (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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. AMC logo.png

Invalid username
Login to AoPS