2004 Cayley Problem 25

The number of positive integers x with x \le 60 such that each of the rational expressions

\frac{7x+1}{2}, \frac{7x+2}{3}, \frac{7x+3}{4}, \cdots, \frac{7x+300}{301}

is in lowest terms (i.e. in each expression, the numerator and denominator have no common factors) is

\textbf{(A)}\ 1\quad \textbf{(B)}\ 2\quad \textbf{(C)}\ 3\quad \textbf{(D)}\ 4\quad \textbf{(E)}\ 5

If there are no answer choices shown, enter a numerical answer.

Full credit to this problem is given to the CEMC, you may view all Cayley contests here.