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2.71828182845904523536……

April 19, 2006 pm30 8:27 pm

So for today a beginning.  We can go a lot further.  The direction is clear.  The end may be less so.

I will add links, decorate, clean up.  All in good time.

I will define some direction, also in good time.  (Think: math, New York City, sewers and trains and streets, teaching).  The direction exists;  it needs to be spelled out.

For now, 2 puzzles:

1. Closed – How many squares are on a regular 8 x 8 checkerboard?  (not just the little ones.  Remember, the whole board is a square, and there are 2 x 2 squares, and 6 x 6 …..)

2. Open – How high can you count with your fingers? (assuming you have 10). How would your counting system work? Is it realistic?  (ring finger and pointer up at once, with the middle finger down, is that realistic?  _|_|_ ? )

Jonathan

6 Comments leave one →
  1. April 30, 2006 pm30 7:11 pm 7:11 pm

    Answer to the counting on your fingers question: Assuming each of my 10 fingers can independently take a recognizably “up” or “down” position, then the answer is equivalent to the highest number represented by a 10-digit binary number: (2^10)=1024.

    As you point out, however, some of these configurations might be difficult or impossible. I just learned that some people do indeed have independent extensor muscles for their middle or ring fingers: http://www.madsci.org/posts/archives/2005-05/1117326759.An.r.html So the answer will vary from individual to individual. But I think that, possibly with some help from the thumb or palm of the hand, produce a recognizable version of “up” or “down” for each finger and could even do the ring/pointer up with others down you propose — just not quickly.

    A deeper investigation into finger-counting might also account for the greater dexterity we have in our index finger. Notice that you can put your index finger “up” and “down” but also “half-up” at the first knuckle. So instead of two possible values, it now has 3. Our total counting ability is now (2^8) (3^2)=2304.

    Of course it’s very difficult to look at someone holding up a collection of fingers in either of these proposed systems and know the number they are indicating. For one thing, the meaning of the left and right hands is different: one is the “high order” bits and the other is the “low order”. People might more commonly encounter this kind of confusion when trying to communicate, say, the number 25. Someone is holding up two fingers on one hand and five on the other… does that mean 25 or 52? Referees in most sports get around this problem by indicating the numbers sequentially: First a two, then a five, for the player wearing twenty five. This system allows them to signal numbers up to 99 in a pretty reliable way, but it requires attention over time, not just a single snapshot of their finger configuration.

    Interesting question!

  2. May 1, 2006 am31 3:23 am 3:23 am

    A computer science professor at the local college wrote one digit on each fingertip, when he held them up he had a 10-digit number.

    When I was in 4th grade, I taught myself to count units on my right hand (0 – 5) and 6’s on my left, bringing me to 35. I still finger count this way.

    I will get this wrong: The Babylonians use a thumb as a pointer to the segments (phalanges) on the same hand, and tracked dozens on the other hand, getting to a round 60 (and providing the basis of the sexigessimal …. I think that I am conflating fact, myth, and fiction.

    That’s the best of the reasonable systems that I can think of. Doesn’t mean there’s not better out there.

  3. May 3, 2006 pm31 3:08 pm 3:08 pm

    You could squeeze a little more than 2^10 by considering hand configuration. Each hand can be presented palm forward or palm backward, giving 2^2 configurations, for a total of 2^12.

    Taking this idea further, it would be easy enough to distinguish four configurations for each hand (palm facing forward, right, backward, left), enabling one to count up to 2^14.

    Each square on the chessboard is uniquely determined by the coordinates of its top-left square. For an n-by-n square, with n = 1, 2, … , 8, there are 9 – n possibilities for each coordinate. So the total number of squares is 1^2 + 2^2 + … + 8^2 = 8*9*17/6 = 204.

    Interestingly, the only positive integers for which 1^2 + … + n^2 is a perfect square are 1 and 24. (See http://mathworld.wolfram.com/CannonballProblem.html)

    (Sorry about the partial comment above. I don’t think your blog software likes a “less than” immediately followed by an “equals” sign!)

  4. May 3, 2006 pm31 7:54 pm 7:54 pm

    <=   I had to try twice to get it, though.  wordpress was suggested to me by a non-math person, and in fact, while I enjoy the puzzles (and have been looking at yours), math is not the primary purpose of this blog (actually, at this point, there are several purposes, none of them primary.  As I gain experience, I might learn more about these sorts of things.

     That being said, I do like your approach to the checkerboard.  It easily extends to "how many rectangles.." whereas the approach I use with students needs to be exchanged for that one.

     jonathan

  5. May 3, 2006 pm31 9:01 pm 9:01 pm

    BTW, thanks for the lead on the packing problem. And I think you have pretty much maxed out the fingers problem, in theory. I don’t think that the “hand-orientations” approach meets the realistic criteria (nor, ro be perfectly honest, does using binary)

    jonathan

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