The number of integers $n$ for which $\frac{1}{7} \leq \frac{6}{n} \leq \frac{1}{4}$ is
(A) 17
(B) 18
(C) 19
(D) 20
(E) 24
Since $\frac{1}{7}$ is positive and $\frac{1}{7} \leq \frac{6}{n}$, then $\frac{6}{n}$ is positive and so $n$ is positive.

Since $\frac{1}{7}=\frac{6}{42}$ and $\frac{1}{4}=\frac{6}{24}$, then the given inequality is equivalent to $\frac{6}{42} \leq \frac{6}{n} \leq \frac{6}{24}$.

Since the fractions are all positive and $n>0$, then this is true when $24 \leq n \leq 42$. (If two fractions have the same numerator, then the larger fraction has a smaller denominator.) Since $n$ is an integer, then there are $42-24+1=19$ possible values for $n$. (We could count the integers from 24 to 42 , inclusive, to confirm this.)

ANSWER: (C)