Jason Ermer has a really cool thing going with Collaborative Mathematics. He wrote to me with an interesting student response that he got, and asked if I’d be interested in sharing it. Good news: I am!

First, here’s the problem:

Then we’ve got the student response.

OK, bad news, I can’t embed this video. Since this is the internet, this may be the point where you’re all like “Oh yeah like I’m going toΒ click on a thing to get to some other interesting thing. Hell, I don’t even know if this other thing is interesting.” Well I’m vouching for this thing. It is interesting. Clicky clicky.

Did you click? Science tells us that some of you didn’t. If you didn’t, then here’s my summary of what the kid said:

“8 is 1000 in binary, so you’ll land on the same finger that you’ll land on when you’re at your eighth finger.”

(See? I told you. I bet now you’re regretting not clicking on the thing.)

Jason is going to write a few posts thinking about this student’s response, but let’s give him a head start on that discussion here.

What strikes you about this kid’s response? Please share any questions that you’ve got about the response. I’m sure we can rope Jason into lurking in the comments.

  • Michael Paul Goldenberg

    Wow. Didn’t think about that at all. Wonder if seeing the previous challenge would have helped, and will check in a bit.

    I did this far less brilliantly, but came to the same conclusion about the finger we should land on. (No, I did NOT count on my fingers all the way to 1000, only to 150 to be sure my idea was correct). What I noticed was that at 25, I was back at my thumb, and that the columns for thumb, forefinger, etc., were changing in interesting ways. The thumb and pinkie always went up by 8. The forefinger and ring finger each went up by 2 then 6, alternately, from whatever their previous number was, so a total of 8 on two passes (or one complete cycle), and the middle finger went up by four on each pass for once again a total of 8 per complete cycle.

    (It seems that 8 makes sense in that you’re ending a cycle when you return to the index finger and starting a new cycle when you get to the thumb, so three fingers in the middle get hit twice in a cycle, while the thumb and pinkie only get hit once each, so it’s as if you had 8 fingers ‘in play,’ not 5 fingers or 10 fingers as you might assume).

    Since 1000 == 0 (mod 8) my first thought was that we’d finish on the thumb, but this was quickly proved wrong when I got to 50 and saw I finished on the forefinger. My guess was that each succeeding group of 25 counts would finish one more finger to the “right” until reaching the pinky, then return to the left and so forth. This pattern held through 150, so I was pretty sure it was correct. At that point, I was ready to finish the problem.

    As there are 1000 = 25 x 40, I concluded that 40 and 1000 would be “congruent” and since 40 lands on the forefinger, so would 1000.

    Pretty inelegant compared with the student’s insight. Now I have to figure out if the student’s answer makes sense and why, and perhaps gain more insight into my own so that I could get it with less counting.

  • I’d love to see some fancy-dancy psychoanalysis of the student that suggests a simple and fascinating reason why she would make this observation. But I’m afraid it something mundane: Mr. Elmer proposes a tricky way of counting, so the student racks her brain for different ways of counting that she’s seen. Not much comes to mind — counting is just counting after all — but then she remembers being taught to count in binary. Could that be what is asked? Wait, we’re counting to 1000, and that’s a number with just ones and zeros as digits, just like in binary! Let me work out what 1000 is when converted back to decimal, and that’s probably the answer.

    One might think the student had some strange insight to produce the answer 8, a number so relevant to the correct to the solution of the problem. But again, I think coincidence is the most likely explanation.

    I’d love to know how she’d have responded if asked to count instead to
    1) 992, which is not a number written in binary
    2) 1011, which I’m willing to bet will lead her to respond 11, which doesn’t make a prominent appearance in a correct solution

    Michael, you stopped too soon when you said 1000 == 0 (mod 8) so we’d finish on the thumb. The process repeats every time you hit your thumb again, and that is every eighth count. But you won’t be on your thumb on the “zeroth count” — if you “back up one” from the starting point, you’ll be on your forefinger. That’s the zeroth position, and your observation indicates that the ending point will be the forefinger.

    Editor’s note: I screwed up the post originally. Jason’s last name is “Ermer,” not “Elmer.” My bad!

  • Michael Paul Goldenberg

    Barry, I kind of suspected that things weren’t quite as simple as 1000 == 0 (mod 8) because of the way this counting was conducted, but the approach I described was just a scrambling first crack at getting an answer. I suspected that it shouldn’t take quite as much grunt work as I put into it, but given the hour and my state of mind, I was pleased anyway. There’s a really nice approach up at the website by Paul Smith that gets at things incisively: http://www.educreations.com/lesson/view/collaborative-mathematics-challenge03-finger-count/12331328/?ref=link

    Would like knowing how he did that video. The math itself is clear.

    Now that I’ve had time to think about the binary approach, I can see that it was one of those truly classic cases we see in math classes: right answer for completely erroneous reasons. To me, examples like this one are worth their weight in gold when trying to show skeptics why expecting students to show their thinking is not only reasonable but essential if we’re going to evaluate their mathematical thinking. Parents get very snide when their child answers a calculation problem with the correct answer and is still expected to explain how it was arrived at. Of course, for elementary arithmetic, it may be hard for students to articulate their thinking beyond saying things like, “I thought about it and knew the answer,” particularly if it’s a matter of recalling facts. But sooner or later, students will face problems where recalling facts will not suffice, or will only appear to do so. This particular example shows a student who did some thinking, applied it, got the right answer, but didn’t test the thinking further to see if it generalized or at least worked for a few other examples. That’s one of the important parts of problem-solving that many students ignore.

  • Hi all, and thanks for the discussion!

    I’ve posted my analysis of this approach over on my own blog. I’d love to hear people’s thoughts, either here or there:
    http://www.collaborativemathematics.org/1/post/2013/11/learning-from-student-mistakes-part-2.html

    Like you, Barry, I’m be super curious to know how the student would respond in other situations. If I hadn’t chosen 1000 as the target number, I doubt the binary numbers connection would ever have come up. I love that a small, random choice could lead to such an interesting approach and discussion. It’s like the Butterfly Effect for math teachers. πŸ™‚

    To your point, Michael G., about “right answer for the wrong reasons”, I wonder whether a video like this could be helpful for students, parents, and others who might ask “why show work?” I wonder how students might respond if I showed this video in the first weeks of class… or to parents on back to school night. πŸ™‚

    When it comes to the modular arithmetic connection, my addition to the conversation is simply to note that we start counting at 1 on the thumb, so any number that is congruent 1 mod 8 will end up on the thumb. We count both 2 and 8 on the index finger, so any number congruent to either 2 or 0 mod 8 (such as 1000) will end up on the index finger.

    I love discussions like this! Thanks for sharing the videos, Michael P.! πŸ™‚

  • A month late, but I’m reading back through the history having only just discovered your site.

    So, as a programmer, I notice that this problem would have been easier to solve if we were all accustomed to counting starting from 0. Counting from 1 to 1000 is the same as counting from 0 to 999, so whereas counting from 1 makes the thumb 1, 9, 19, 25 (whose closed form is not immediately obvious without thought), counting from 0 makes the thumb 0, 8, 16, 24 (which is quite obvious). Every other finger is then the thumb’s value, plus how many spaces away from the thumb it is. That’s the same in both cases, but because it’s easier to see when counting from 0, it makes us more confident in our solution. [1]

    But while this student’s answer is random luck, it’s interesting to see why. 1000 in binary is correctly determined to be 8 in decimal, but it’s also 10 in octal (base 8). On the other hand (if you’ll excuse the pun) 1000dec = 1111101000bin, which is 1750oct [2]. And, as 1750oct and 10oct end in 0, it’s really just a round-about way of saying 8dec and 1000dec are both divisible by 8dec. Or, it’s a whole number of “hand counting passes”, thus ending on the index finger.

    And we can extend it further: if counting from 1, any octal number ending in:
    1 will be the thumb.
    0 or 2: index finger
    3 or 7: middle finger
    4 or 6: ring finger
    5: pinky finger

    Though I personally think it has better symmetry when counting from 0, letting you see the “loop back” in the numbers themselves:
    0: thumb
    1 or 7: index finger
    2 or 6: middle finger
    3 or 5: ring finger
    4: pinky finger

    [1] I’ll let Edsgar Dijkstra explain all of the reasons why counting from 0 is superior to counting from 1: http://www.cs.utexas.edu/users/EWD/transcriptions/EWD08xx/EWD831.html

    [2] Binary numbers are quite easy to convert to octal numbers. You can partition the binary value into groups of three digits (starting from the right) and figure out the octal digit for each group. 1111101000bin -> (XX1) (111) (101) (000) -> (1) (7) (5) (0) => 1750oct. You can’t do that with decimal to binary.