# Problem of the Week Problem D and Solution Count on This

## Problem

Determine the number of integer values of $$n$$ that satisfy the following inequality: $\dfrac{1}{9} \leq \dfrac{7}{n} \leq \dfrac{1}{5}$

## Solution

First notice that since $$\tfrac{1}{9} \leq \tfrac{7}{n}$$, and $$\frac{1}{9}$$ is positive, that means $$\frac{7}{n}$$ must be positive as well. It follows that $$n$$ is positive.

Since $$\frac{1}{9}=\frac{7}{63}$$ and $$\frac{1}{5}=\frac{7}{35}$$, we can rewrite our inequality as follows: $\frac{7}{63} \leq \frac{7}{n} \leq \frac{7}{35}$ Since the fractions are all positive and $$n>0$$, this is true when $$35 \leq n \leq 63$$. This is because if two fractions have the same numerator, then the larger fraction must have a smaller denominator, i.e. $$\frac{2}{5}<\frac{2}{3}$$.

Now we just need to count the number of values of $$n$$ that satisfy $$35 \leq n \leq 63$$. We could count them, but a faster way would be to do some simple math. Since $$n$$ is an integer, there are $$63-35+1=29$$ possible values for $$n$$.