A Little Light Algebra

This is the kind of thing  I really enjoy.

Nrich have also made a simpler version with only two  lights and more guidance on how to understand and solve the patterns. That said, as the second version is explicit about how to represent the patterns algebraically, I suspect it is more likely to alienate or ‘freeze’ pupils who are perfectly able to identify and solve the patterns but have preconceptions about algebra and their abilities. I don’t think this is an error on the part of nrich – more a caveat when showing it to students.

I haven’t tried it in class yet – I’m trying to decide if it’s better for introducing or consolidating patterns/nth term, or just as a stand-alone piece for a one-off lesson.

PS – I was able to find a number when all four would light up, but only by trial and improvement (in the case below, only using multiples of 11n-3), not by setting it to equal zero. I’m not sure if I’m missing an obvious method, but I couldn’t get everything ‘to one side’ without new things popping up elsewhere! How should one go about solving an ‘equation’ such as this for one value of n:

8n-6 = 11n-3 = 3n-1 = 6n-4 ?

UPDATE: I’ve realised they can’t all have the same value of n when they’re solved, as it’s the number generated (74, in this case), not the number of the term (10, 7, 25, and 13, respectively), that they have in common. I’m flummoxed now!

3 Comments

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3 responses to “A Little Light Algebra

  1. I’ve been watching Oz lately, and in a recent episode, a load of the prisoners go back to school to get their high school diploma.

    This blog’ll get me the maths education my inner 12 year old never had…

  2. Sadly, my solution to this puzzle isn’t very elegant, but it works. Unfortunately it requires modular arithmetic (which isn’t really appropriate for school kids, unless you do base 12). And there aren’t enough equations to do simultaneous equations properly (which you can tell instantly by the lack of a unique solution).

    Anyway, if we quickly re-label your unknowns, so that we want to solve:
    x = 8a-6 = 11b-3 = 3c-1 = 6d-4
    Then we can get that

    x = 2 (mod 8) (1)
    x = 8 (mod 11) (2)
    x = 2 (mod 3) (3)
    x = 2 (mod 6) (4)

    We need to find values of x that satisfy all of these, and we proceed with pairs of equations, starting with (3) and (4) because 3 and 6 have the easiest common multiple.

    So (3) becomes x = 2,5 (mod 6), so clearly to be consistent, we must take that x = 2 (mod 6), so equation 3 is redundant.

    Next, we take (4) and (1) to get that:
    x = 2,8,14,20 (mod 24) and
    x = 2,10,18 (mod 24)
    so clearly x = 2 (mod 24) (5)

    lastly, we combine (5) and (2) with base 11*24 = 264

    x = 2,26,50,74,98,122,146,170,194,218,242 (mod 264)
    x = 8,19,30,41,52,63,74,85,96,107,118,129,140,151,162,173,184,195,206, 217,228,239,250,261 (mod 264)

    so the solutions are x = 74 (mod 264)

    Hopefully this helps you to understand it, even if it’s not going to be hugely useful in school.

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