Wednesday, June 5, 2013

Number Patterns in Problem Solving.

This appeared on my Facebook page today. I get lots of these with the invitation to solve it. I'm not very good at FB so I never like to click on anything I think is going to get me into trouble. Who knows what is lurking the other side of a smple mouse click?

Anyway this is a really neat little problem that very eloquently demonstrates the power of pattern in learning math. As I have said many times before the identification of a numercial pattern changes everything in terms of trying to remember facts as well as more higher order thinking activities such as genuine problem solving. Most algebraic problems require the application of a pattern of some sort.

So what does the idea of pattern have to do with this little FB treat? First there is a pattern to each of the descending series of numbers; at least until the 3 is reached. They descend by 1 eachg time. The numbers on the right also descend but by different amounts, 14, 12, 10. extending these patterns, if there was a 4 on the left the number on the right would be 12 ( 8 less than 20), then would come 3 on the left and 6 on the right (6 less than 12). Now look at th relationship between each of the pairs of numbers and you get 8x7=56. 7x6 = 42, 6x5=30, 5x4=20, 4x3=12, and 3x2=6. Each number descends by 1. I could be wrong, of course, and completely missed the trick, if there is one, to get a differerent answer.

I'm not sure if one needs to be a genius to solve this. I would expect any 4th grade student to be able to solve this if they have been taught that math is the science of pattern and not a random collection of facts and ideas.



6 comments:

  1. A general solution is posted here . . . https://www.scribd.com/doc/260182194/Elementary-Sequences

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  2. 8=56 56-42=14 8x(7), 7x2=14
    7=42 42-30=12 7x(6), 6x2=12
    6=30 30-20=10 6x(5), 5x2=10
    5=20 always -2 20-12=8 5x(4), 4x2=8
    3=? this just flip 4x3 and 3x4 to get different answer

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  3. It turns out the ‘?’ can be whatever we want it to be!

    Allow me to illustrate. Take for example, f(x)=(1/40)x^4-(13/20)x^3+(291/40)x^2-(553/20)x+42, and evaluate f(8), f(7), f(6), f(5), f(3). It turns out that:

    f(8)=56,
    f(7)=42,
    f(6)=30,
    f(5)=20,
    f(3)=9.

    Wait, what sorcery is this? Turns out that although the popular rule f(x)=x(x-1), which gives f(x)=6, satisfies the known values in the sequence, that f(x)=(1/40)x^4-(13/20)x^3+(291/40)x^2-(553/20)x+42 also satisfies them–except with a different value of f(3)!

    Here’s another one that also works but gives f(3)=12:
    f(x)=(1/20)x^4-(13/10)x^3+(271/20)x^2-(543/10)x+84

    And here is one where f(3)=π
    f(x)=(1/120)(π-6)x^4-(13/60)(π-6)x^3+(1/120)(251π-1386)x^2+(1/60)(3138-533π)x+14(π-6)

    Finally, in general if you want the ‘?’=k, i.e., f(3)=k where k is the value of your choice, then
    (1/120)(k-6)x^4-(13/60)(k-6)x^3+(1/120)(251k-1386)x^2+(1/60)(3138-533k)x+14(k-6)

    More details here: https://www.scribd.com/doc/260182194/Elementary-Sequences
    For a non-polynomial rule see here: http://i.imgur.com/BHkg0Ad.png

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