At a party once a guy bragged that he could count up to x on one hand.
“Nah-uh!” we all protested “you’ve only got five fingers. How could you possible count as high as x?”
He then put on a strange show of flipping up various fingers then retracting them as he counted up to x. The idea was that each finger did not represent a thing to be counted (one finger equals 1, two fingers equals 2, three fingers equals 3, four fingers equals 4, five fingers equals 5, and finished), but rather it was the number of combinations of fingers that could be extended that leads us up to x.
I’ve been trying to remember how he did it, but the problem is I don’t remember was x was.
So far I’ve come up with 23 (24 if you count having zero fingers extended- should you?).
But I think the guy at the party got higher. Try as I may, I can’t get any higher than 23- but it may be that I just confuse myself in the process (sometimes I can’t get to 23), so I come to the SDMB for help.
He explained the combinations as representing binary code. If a finger was not extended it was a zero, if it was extended it was a 1. So this would be easy to figure out if you know anything about binary, or at least how to calculate the possible variations of having two possible figures within five total digits. Unfortunately, I don’t know any of this (in fact the true math experts of the SDMB may be lost in my layman’s gibberish).
Anyway, here are the 23 that I came up with (displayed as each digit representing a finger, thumb on the right and pinky on the left. An extended finger is a 1, a nonextended finger is a 0):
Also, the only order I’ve got here is just what I’ve come up with to try to help myself to not repeat any. I don’t actually know the correct order that these numbers should be in. If you could help with that, it would be appreciated.
How high can you count using both hands???