Review question

# For how many integers $n$ is $n/(100-n)$ also an integer? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R8880

## Solution

For how many integers $n$ is $\dfrac{n}{100-n}$ also an integer?

#### Method 1

Let $\dfrac{n}{100-n}=x$ where $x$ is an integer. Hence $n=100x-nx$ (so $x$ cannot be -1).

Adding $nx$ to both sides of the equation and factorising, we have $n(1+x)=100x$, giving $n=\dfrac{100x}{1+x}$.

Now $x$ and $1+x$ can have no common factors greater than $1$. Therefore, as $n$ is an integer, $1+x$ must be some factor of $100$.

Hence, $x+1$ can take any of the values $\pm1,\pm2,\pm4,\pm5,\pm10,\pm20,\pm25,\pm50,\pm100$.

This gives $18$ different values for $x+1$.

Each value of $x+1$ gives unique corresponding values of $x$ and $n$, so there are $18$ possible values of $n$ for which $\dfrac{n}{100-n}$ is an integer.

#### Method 2

Let $n = k + 100$. We see that $n$ is an integer if, and only if, $k$ is an integer.

So let’s replace $n$ with $100 + k$ in $\dfrac{100}{100-n}$, giving $\dfrac{100+k}{k} = 1 + \dfrac{100}{k}.$

Now this is an integer if, and only if, the integer $k$ divides into $100$ exactly. There are $18$ integers that do this.