AMC12 2003 A

For any positive integer $n$ which is not a perfect square, exactly half of its positive factors will be less than $\sqrt{n}$, since each such factor can be paired with one that is larger than $\sqrt{n}$. (By contrast, if $n$ is a perfect square, one of its factors will be exactly $\sqrt{n}$, which would therefore have to be paired with itself.) Since $60$ is indeed not a perfect square, it follows that half of its positive factors are less than $\sqrt{60} \approx 7.746$. This estimate clearly shows that there are not even any integers, let alone factors of $60$, between $7$ and $\sqrt{60}$. Accordingly, exactly half of the positive factors of $60$ are in fact less than $7$, so the answer is precisely $\boxed{\mathrm{(E)}\ \frac{1}{2}}$. Testing all positive integers less than $7$, we find that $1$, $2$, $3$, $4$, $5$, and $6$ all divide $60$. The prime factorization of $60$ is $2^2 \cdot 3 \cdot 5$, so using the standard formula for the number of divisors, the total number of divisors of $60$ is $3 \cdot 2 \cdot 2 = 12$. Therefore, the required probability is $\frac{6}{12} = \boxed{\mathrm{(E)}\ \frac{1}{2}}$. Though this is not recommended for reasons of time, one can simply write out all the factors of $60$, eventually finding that \[60 = 1 \cdot 60 = 2 \cdot 30 = 3 \cdot 20 = 4 \cdot 15 = 5 \cdot 12 = 6 \cdot 10.\] Hence $60$ has $12$ factors, of which $6$ are less than $7$ (namely, $1$, $2$, $3$, $4$, $5$, and $6$), so the answer is $\frac{6}{12} = \boxed{\mathrm{(E)}\ \frac{1}{2}}$.