Ah a little down one of the side paths of amateur radio for this question from the General Class exam pool.

## How many states does a 3-bit binary counter have?

A. 3

B. 6

C. 8

D. 16

The answer depends on us knowing how to count in binary. Essentially the question is asking us what the maximum number of an 3-bit integer is. Your first instinct is probably to yell out “well 3 of course!”

But you’d be wrong. Let’s look at why.

In a 3-bit counter, there are (wait for it) three places that can either be a 1 or a 0.

So, for example. 010, or 100, or 111, or 001… It should start to make sense.

Here’s the pro-tip of the day. The maximum number of states in a binary system depends solely on the number of bits, and the answer is always 2^{n} where n is the number of bits.

So, ergo qed and all that nonsense, the answer is 2^{3}. Or 8. 1*2 is 2, 2*2 is 4, and 4*2 is 8.

**The answer is C) 8. **

Now I know what you’re thinking. “What does this have to do with amateur radio?” And on the surface its a valid question. But when you start getting into digital modes and encodings, bit rates and how much information can be sent in a given time period with a given signal type… This starts to matter.

Strictly speaking, though, its not likely to come up in day-to-day amateur radio operations. They just want you to be prepared.

Here’s the table of all the possible states of a 3-bit integer:

000 | 0 |

001 | 1 |

010 | 2 |

011 | 3 |

100 | 4 |

101 | 5 |

110 | 6 |

111 | 7 |

[G7B05]