I made these mistakes, really just minutes ago, in attempting to solve a problem from brilliant.org.

my math mistake

 

Here was my written work:

IMG_2902

 

What was I thinking? What did I understand? What didn’t I understand? What questions would you ask to help me?

[Also, my ego does require me to announce that I got this problem on the third try. Maybe one of these days I’ll post something that I’m still stuck on.]

[Also also, do you like analyzing this sort of work? The problem is kind of tougher than what usually gets posted around here, so that I imagine that a lot of folks will skip this post once they don’t immediately see the answer to the question. But, hey, I’m curious to see what you all come up with. The advantage is that I can verify or disconfirm any theories that you might have about what the “student” was thinking.]

  • not 100% sure where the 15 comes from…and i may be completely off-base, but is it 498? My guess is you were thinking about a way to represent what would be an enormously long list of prime factors of 1000!, then count the 3s? so there are 333 factors with at least a 3, 111 with a 9 (or two threes, but you already counted one of them…), 37 with a 27 (three 3s, but you already used two of them…) etc. How I would help? Without being 100% sure of my answer, I wouldn’t want to lead you astray but I’d just talk my way through it – because I’d assume that a student getting this type of problem would be able to have a pretty logical math discussion. Anyway, long story short: 498?

  • I came up with the same answer as Mr. T. My work is here: http://mrkraft.wikispaces.com/file/view/SAM_0714.JPG/440176204/SAM_0714.JPG
    I’m very curious to know how you and he figured this out.

    I’m also really curious about is “1+2+3+4+5=15”. Was this an attempt to find the number of 3^6 integers, 3^5 integers, 3^4 integers, and so on? I’m not sure.

    I suppose I would push the student to specifically tell me where 3^333 and 3^15 come from? Which integers does 3^15 represent? Are we sure we didn’t miss something?

  • Michael
    Looking at your work it seems that you are making the mistake of concluding that, since 999/3 = 333 that there must be 333 instances of the number 3 built into that. Sliding from divisibility ideas into exponent/factoring ideas is a very easy and natural mistake to make and I think it’s instructive to see teachers making this kind of mistake as well. A quick way to point out this mistake is to simply start with your list above of the first ten integers and see that the 9 creates 2, not 3, instances of a 3 as a factor. This initial setup of yours reminds me of the best student work I’ve seen in solving the classic locker problem (http://connectedmath.msu.edu/CD/Grade6/Locker/)
    I would ask you to slowly expand your thinking from the 10! list into a 20! list next and see if/when we can make the leap all the way to our goal of 1000!

  • Laurel Pollard

    I worked on this at lunch yesterday with my spouse. I think I would ask you about your first strategy (solving a simpler problem) and ask you to explain what you noticed. Could you express 9 = 3×3 differently or explain why? Would it be helpful to think of all the 3’s as powers. Such as 3 as 3^1 and if so would that help you with making a organized list in your second strategy 999 / 3^1; 999/3^2, 999/ 3^3, etc. which is what I noticed in the middle of your work.

    I wonder what this says about factorials?

  • It looks like you first counted all the numbers divisible by 3, which is a good start. Then you realized that 9, 27, 81, 243, and 729 would have more factors of 3 in them, and counted the extras there. So, the next step is to have you look at a number like 18 (maybe by asking you to write out *all* the prime factors in 20!) so that you can discover that you need all the multiples of 9, 27, 81, and so on, as well.

  • I’m going with 3^555. I think I’m over but I’m going for it.
    I’d ask you to explain your thought process in dividing 999 by specific numbers. Why did you choose the numbers that you did? Did you account for every multiple of 3 and 9? What about factors of 999 that have additional factors besides 3 or 9? This one hurt my brain. Thanks. I needed that.