A bandwidth of 6 MHz is available for A.M. transmission. If the maximum audio signal frequency used for modulating the carrier wave is not to exceed 6 kHz. The number of stations that can be broadcasted within this band simultaneously without interfering with each other will be _________.

We begin by recalling the basic fact about ordinary amplitude-modulated (A.M.) radio transmission: when a carrier of frequency $$f_c$$ is modulated by an audio (message) signal whose highest frequency component is $$f_m$$, two side-bands are produced, one above and one below the carrier. The upper side-band extends up to $$f_c + f_m$$, while the lower side-band goes down to $$f_c - f_m$$. Thus the total spectral span occupied by that single station is twice the highest modulating frequency.

Mathematically, the required bandwidth for one A.M. station is stated as

$$\text{Bandwidth per station} = 2\,f_m.$$

Now the numerical values given in the problem are:

Maximum audio (modulating) frequency: $$f_m = 6\ \text{kHz}.$$

Therefore, using the above formula, the bandwidth demanded by one station is

$$\text{Bandwidth per station} = 2 \times 6\ \text{kHz} = 12\ \text{kHz}.$$

The spectrum segment reserved for all the A.M. broadcasts together is stated to be

$$\text{Total available bandwidth} = 6\ \text{MHz}.$$

Since $$1\ \text{MHz} = 1000\ \text{kHz}$$, we convert this to kilohertz to match units:

$$6\ \text{MHz} = 6 \times 1000\ \text{kHz} = 6000\ \text{kHz}.$$

We now find out how many non-overlapping 12 kHz slots can be fitted into 6000 kHz. The number of simultaneous stations, $$N$$, is simply the ratio of the total available bandwidth to the bandwidth per station:

$$N = \frac{\text{Total bandwidth}}{\text{Bandwidth per station}} = \frac{6000\ \text{kHz}}{12\ \text{kHz}}.$$

Carrying out the division, we get

$$N = \frac{6000}{12} = 500.$$

So, the answer is $$500$$.