The society on Jupiter's moon Io has a very odd way of counting. They count in binary using special counters: small discs with 0 printed on one side and 1 on the other. To increase speed when counting they made a rule:(adsbygoogle = window.adsbygoogle || []).push({});

Rule #1: To get the next number in the sequence, only one counter can be turned over.

To ensure all of their numbers are unique they made a second rule:

Rule #2: The next number in the sequence must not have already appeared.

They start counting with a single counter showing 0 and then flip it to get 1. To progress any further a new counter is needed. Again, to ensure speed, they made this third rule:

Rule #3: When no more numbers can be made a single counter can be added on the left.

They can't put the new counter down with the 0 on top to get 01 because 01 equals 1, which has already been used. Instead, the sequence continues like this:

11, 10

Now another counter must be added on the left:

110

Here there appears to be a choice: do they flip the second counter along to get 100 or the third to get 111? To avoid ambiguity they made a fourth rule:

Rule #4: The next number in the sequence must have the lowest possible value.

So the sequence continues like this:

100, 101, 111, 1111, 1011, 1001, 1000, 1010, 1110, 1100, 1101, 11101, 10101, etc.

The question is: can the Ionians continue to count like this indefinitely while adhering strictly to their four rules?

If you solve the puzzle please explain how you did it.

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# I Mathematics binary puzzle

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