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.