Compute the minimum number of 120Ω resistors to get 80Ω of resistance?
A direct solution can be found through the application of continued fractions.

If what you have is 120&Omega; and what you want is 80&Omega;, write down the fraction:

$$\frac{80\Omega}{120\Omega} = 0.6667$$

Since the integer part is zero, you'll be starting by putting resistors in parallel. Invert the fractional part:

$$\frac{1}{0.6667} = 1.5$$

This tells you that you'll have 1 resistor in parallel with some number of resistors in series. Invert the fractional part again:

$$\frac{1}{0.5} = 2.0$$

This tells you that you need 2 resistors in series. Since there is no fractional part at this point, you're done.

The answer is a total of 3 resistors.