In her solution, Maile observes that the numeral 1000 can represent different numbers. It might represent

*one thousand*. In fact, I think it's fair to say it "usually" represents one thousand because we usually interpret numerals in base 10. But, the symbol 1000 could just as easily represent

*eight*if we interpret it in base 2. (In fact, it could just as easily represent a lot of numbers:

*sixty-four*, f

*our thousand ninety-six*,

*negative twenty-seven*... but let's not get ahead of ourselves.)

Maile answers the challenge by claiming that we can stop finger counting at eight rather than counting all the way to one thousand, since 1000 represents the number eight in base 2. She's right in that we do in fact land on the same finger. "But," I wondered, "Does that always work?"

(As a quick aside, I have to appreciate the lucky randomness of the fact that I chose 1000 as the target number. That is to say, if I had asked what finger we'd be on when counting to 500 or to 2000, the connection to binary numbers would never, I suspect, have come up.)

To answer the question "does it always work?", it's helpful to watch Paul's solution to the challenge, in which he makes the connection between this kind of finger counting and modular arithmetic. If a number is congruent to either 2 or 0 modulo 8 it will end up on your index finger. Since \(1000 \equiv 0 \mod{8}\), we end up on the index finger when we count to 1000.

Said the other way around: if two numbers are congruent modulo 8, then they will end up on the same finger when finger counting in this way. So here was the question I had:

Given a numeral made up only of the digits 1 and 0, are the base 2 value of this numeral and the base 10 value of this numeral always congruent modulo 8?