The equivalent resistance of series combination of two resistors is $$s$$. When they are connected in parallel, the equivalent resistance is $$p$$. If $$s = np$$, then the minimum value for $$n$$ is ________. (Round off to the Nearest Integer)

Let the two resistors have resistances $$R_1$$ and $$R_2$$. Their series combination gives $$s = R_1 + R_2$$, and their parallel combination gives $$p = \frac{R_1 R_2}{R_1 + R_2}$$.

We are given that $$s = np$$, so $$R_1 + R_2 = n \cdot \frac{R_1 R_2}{R_1 + R_2}$$, which gives $$(R_1 + R_2)^2 = n \cdot R_1 R_2$$.

We know from the AM-GM inequality that for positive real numbers, $$R_1 + R_2 \geq 2\sqrt{R_1 R_2}$$. Squaring both sides, $$(R_1 + R_2)^2 \geq 4 R_1 R_2$$.

Since $$(R_1 + R_2)^2 = n \cdot R_1 R_2$$, we get $$n \cdot R_1 R_2 \geq 4 R_1 R_2$$, which simplifies to $$n \geq 4$$.

The minimum value of $$n = 4$$ is achieved when $$R_1 = R_2$$ (equality condition of AM-GM). We can verify: if $$R_1 = R_2 = R$$, then $$s = 2R$$ and $$p = \frac{R}{2}$$, so $$\frac{s}{p} = \frac{2R}{R/2} = 4$$.

Therefore, the minimum value of $$n$$ is $$\boxed{4}$$.