Diophantine Resistors

We know that to get a resultant resistance lower than the original component resistance of $1000 \ \Omega$ , the connection must be in parallel. Let the equivalent circuit of $600 \ \Omega$ with the least number of resistors in parallel which is two. Let one of the resistor be $1000 \ \Omega$ , then the other resistor $R_1$ is given by:

$\begin{aligned} R_1||1000 & = 600 \\ \frac {1000R_1}{1000 + R_1} & = 600 \\ 1000R_1 & = 600000 + 600R_1 \\ R_1 & = 1500 \ \Omega \end{aligned}$

We know that the resultant resistance of two similar resistors in parallel is half the resistance of one of the resistor. Therefore, we have:

$\begin{aligned} 1500 & = 1000 + 500 \\ & = 1000 + 1000||1000 \end{aligned}$

That is, one $1000$ - $\Omega$ resistor in series with two $1000$ - $\Omega$ resistors in parallel. And the final equivalent circuit is these three $1000$ - $\Omega$ resistors in parallel with another single $1000$ - $\Omega$ resistor, therefore, a total of $\boxed{4}$ $1000$ - $\Omega$ resistors.

The LCM of $600$ and $1000$ is $3000$ .

Assuming that it has to be a parallel network (since the resultant resistance is smaller than the original resistance/s).

We get that the resultant should have been of this form

$\frac{1}{R} = \frac{5}{3000}$ .

If we investigate the partitions of $5$ providing us with resistances which are easily achievable, we get this

$\frac{2}{3000} + \frac{3}{3000}$

Which means to achieve $600 \Omega$ , we need $1500 \Omega$ and $1000 \Omega$ in parallel.

To achieve $1500 \Omega$ , we find that we can split it into $1000 \Omega$ and $500 \Omega$ , with which the latter can be achieved by having two $1000 \Omega$ resistors in parallel.

Thus, we can just need four resistors to achieve $600 \Omega$ .

TL;DR version:

$(1000 + (1000||1000))||(1000)$

$(1000 + 500)| (1000)$

$(1500)||(1000)$

$600 \Omega$

Let 1000 ohm = R Required Resistance is 600 i.e. 3R/4

Therefore, it is clear that parallel circuit is involved (Resultant < Each Resistor)

Formula for R1 and R2 in parallel:

$\frac{R_{1} X R_{2}}{R_{1} + R_{2}}$

therefore by inspection one can see that R1 X R2 =3R*R and R1 + R2 = 3R+R

Hence, 4 resistors are required

3 in series and 1 in parallel