What is the least number of 8 Ω resistors you can arrange to have a total resistance of 5 Ω and how do you do it?
I never faced this particular question before. But faced with it now, I'd start out (after mentally verifying that a bridge arrangement would be pointless -- and it is) by asking myself what, in parallel with \$8\:\Omega\$ would yield \$5\:\Omega\$?

It turns out that the answer is \$13\frac{1}{3}\:\Omega\$.

Then I'd wonder about what might make that. Well, if I had \$8\:\Omega\$ already, then I'd need another \$5\frac{1}{3}\:\Omega\$ to get that total. Well, luckily \$16\:\Omega\vert\vert 8\:\Omega\$ makes that.

So that's my answer. It's not a general purpose algorithm to get from \$X\$ to \$Y\$, exactly. But it's a thought process to find an answer here. And it suggests an algorithm (discussed below.)

The answer I'd give is:

$$\left(\left(\left(8\:\Omega+8\:\Omega\right)~\vert\vert~8\:\Omega\right)+8\:\Omega\right)\vert\vert ~8\:\Omega$$

or,







simulate this circuit &ndash; Schematic created using CircuitLab

The algorithm might be to realize that 8 and 5 are relatively prime and that it is likely that you'll need to reach their product, \$8\cdot 5=40\$, in order to find an answer. Intuitively, this makes some sense thinking that you are probably looking for a least common multiple of the two values. So it does strongly suggest that 5 resistors will be the minimum you can use here.