An association of 4 resistors of 30ohms each, how can one assemble them so you get a resistance equivalent of 18ohms?
Clearly to get 18 ohms, you need some of the resistors in parallel.
So, what do you need in parallel with a 30 ohm resistor to get 18? Answer: 45 ohms.
Now, you have a simpler problem: how to make 45 ohms from three 30 ohm resistors. That should be obvious!
Some of the earlier answers have already "given you a fish," but "teaching you how to fish" is more useful in the long term IMO.
There is a technique I once found in a graph theory book by Béla Bollobás of all places. Imagine having a resistor network where the schematic can be draw without any crossings. Then for a given potential applied to the network, measure the potential and current for each component resistor. If you replace each resistor with a thin resistive plate V units tall and I units wide (V=IR so the aspect ratio is the resistance), the rectangles assemble into a single large rectangle whose aspect ratio is the equivalent resistance, and each junction shows up as a horizontal line.
This suggests a way to look for a resistor network: take four rectangles 30 units tall and 1 unit wide, and we want to find a way to scale the rectangles and assemble them into a rectangle that is 18 times taller than it is wide.
For easier visualization, since all the resistors are equal we can scale everything down vertically: the problem is to make a 18:30 (that is, 3:5) rectangle out of four squares. For illustration, I found I think every possible rectangle size and resistance you can make with these four resistors, but knowing 3:5 is the target size can speed up the search.
(Edit: I forgot 1:4, giving 7.5 and 120 Ohms.)
The equivalent resistance is 30 times the rectangle's aspect ratio. There's no need to do any sort of reduction of series/parallel circuits. An interesting bonus is that by rotating the rectangle, you get a circuit whose resistance is the reciprocal times 30 of the original circuit.
The rectangle method can be somewhat useful in the case of a network that can't be reduced using the series/parallel equations, though I think this one can be done by a delta-wye transformation:
Put two in parallel, trim them resistors to 18R, then throw the rest in your parts bin for later.
Or, assuming they're 10% resistors, find the ones that are on the low end (27R) and put two in series, putting that in parallel with one 27R. Put the last one in your parts bin for later.
take the four 30 ohms resistors back the shop and ask for an 18 ohm resistor
- you might get some money back
- it is quicker to wire up
- is will cost less to produce
- it looks neater
- it will be more reliable
Invert the problem
Instead of seeking a resistor of 18 ohms, seek a conductor whose conductance is 1/18 siemens.
You have four conductors of 1/30 siemens.
When you parallel conductors, their conductance simply adds.
Placing conductors in series requires some fairly tricky math**, so invert them back into resistors for that math, because series resistances simply add.
** 1 / (1/C1 + 1/C2) ... Look familiar?