An npn transistor operates as a common emitter amplifier with a power gain of $$10^6$$. The input circuit resistance is 100 $$\Omega$$ and the output load resistance is 10 k$$\Omega$$. The common emitter current gain $$\beta$$ will be (Round off to the Nearest Integer)

In a common emitter transistor amplifier, the current gain $$\beta$$ is defined as the ratio of the output (collector) current to the input (base) current: $$\beta = \dfrac{I_C}{I_B}$$.

The voltage gain is $$A_V = \beta \times \dfrac{R_L}{R_{in}}$$, where $$R_L$$ is the output load resistance and $$R_{in}$$ is the input resistance. The current gain of the amplifier is $$A_I = \beta$$. The power gain is the product of voltage gain and current gain: $$A_P = A_V \times A_I = \beta \times \dfrac{R_L}{R_{in}} \times \beta = \beta^2 \times \dfrac{R_L}{R_{in}}$$.

We are given: $$A_P = 10^6$$, $$R_{in} = 100 \; \Omega$$, and $$R_L = 10 \text{ k}\Omega = 10{,}000 \; \Omega$$.

Substituting into the power gain formula: $$10^6 = \beta^2 \times \dfrac{10{,}000}{100} = \beta^2 \times 100$$.

Solving for $$\beta$$: $$\beta^2 = \dfrac{10^6}{100} = 10^4$$, so $$\beta = \sqrt{10^4} = 100$$.

The answer is $$\boxed{100}$$.