April 2nd, 2014, 06:22 AM  #1 
Math Team Joined: Apr 2012 Posts: 1,579 Thanks: 22  5th Grade Number Theory Time Again!
I will once again soon be visiting a friend's fifth grade class to get them all abuzz about math as best I can. Last time, we did a fair amount of factoring tricks and a little on number sequences. I want to do more on number sequences for some prominent classes of figurate numbers, with special focus on how they interrelate. Of course, we'll do triangular numbers and squares, how they interrelate, how each can be derived from serial addition (all numbers for triangular numbers and odds for squares). I will then "dissolve" triangular numbers into four smaller triangular numbers of a special sort via the various interrelations. I will move to the "3D extensions" of triangular numbers and squares, ie tetrahedral and square pyramidal numbers and how they interrelate via triangular dipyramidal numbers. This is likely more than enough, especially since we will be building these numbers with styrofoam balls so that the kids can FEEL how they work. But if anyone has some suggestions for further adventures, let me know! 
April 2nd, 2014, 06:34 AM  #2 
Math Team Joined: Apr 2012 Posts: 1,579 Thanks: 22 
Nice typo in the subject line, no?

April 2nd, 2014, 08:35 AM  #3 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1038  
April 2nd, 2014, 08:56 AM  #4 
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms  I've always been amused by the idea that if you can fit figurate numbers to a polynomial of the appropriate degree, then it's the right one. I don't know if you can work that in? At a higher level this would lead to Bernoulli numbers and the Faulhaber formula.

April 2nd, 2014, 09:13 AM  #5  
Math Team Joined: Apr 2012 Posts: 1,579 Thanks: 22  Quote:
Two things I hope to touch on beyond what I mentioned are my own observation that the finite diagonals on the multiplication table add up to tetrahedral numbers. MMF member icemanfan gave a proof. I will just lead the kids to spot the pattern. Second thing will be the known thing about square pyramidal numbers defining how many squares are in a n x n square grid. Again, I'll let them count up a few and form their own conjecture once they know the number sequences and leave it as something for those especially interested in to look into further.  
April 2nd, 2014, 09:59 AM  #6 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs  
April 2nd, 2014, 10:03 AM  #7 
Math Team Joined: Apr 2012 Posts: 1,579 Thanks: 22  
April 2nd, 2014, 10:29 AM  #8 
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms  OK. A polygonal number clearly must grow as (some constant) * (some linear measure)^2 + (lowerorder terms) since it describes a twodimensional shape. (If you scale it up by a factor of 2 in all linear dimensions, its area must quadruple.) So if you could describe the number of points in a kpolygonal figure by a polynomial, it would have to be quadratic. This is 'obvious' (though not to 5th graders, I imagine!). But for any k, you can represent it as a quadratic polynomial. This is not obvious at all! But knowing that it is a quadratic, you might not  should not  be satisfied. Which one? The nth square is n^2, clearly, and you may know that the nth triangular number is n(n+1)/2. But what is the formula for a pentagonal number, for example? Well, the smallest pentagonal number is n = 1 which has only a single point, and the next one at n = 2 has 5 points, naturally enough. With a bit of effort you can see that the third pentagon has 12 points, or you can use a trick: the 0th pentagonal number is 0. Either way you have three points. Let's say the equation for the nth pentagonal number is a*n^2 + b*n + c for some numbers  we don't know which!  a, b, and c. Then we need a*n^2 + b*n + c to be 0 when n is 0, so a*0^2 + b*0 + c = 0 We also need a*1^2 + b*1 + c = 1 and a*2^2 + b*2 + c = 5 and these three can be simplified to c = 0 a + b + c = 1 4a + 2b + c = 5 Knowing that c is 0 we can get a + b = 1 4a + 2b = 5 and a bit of magic subtraction 4a + 2b  (a + b) = 5  (a + b) 4a + 2b  (a + b)  (a + b) = 5  (a + b)  (a + b) 3a + b  (a + b) = 5  (a + b)  (a + b) 2a = 5  (a + b)  (a + b) But we know that a + b = 1, so this is 2a = 5  1  1 = 3 and this gives a = 3/2. Then since a + b = 1 we have 3/2 + b = 2 and so b = 1/2 and we get the equation an^2 + bn + c is just (3/2)n^2 + (1/2)n which gives the nth pentagonal number (modulo mistakes!). The same is true of higherdimensional shapes as well. But I don't think I'd work through this with them, just mention that it is possible. If they can find a polynomial that works on three points (by trial and error, say), then it works for all possible choices. Amazing! Last edited by CRGreathouse; April 2nd, 2014 at 10:40 AM. 
April 2nd, 2014, 12:15 PM  #9 
Math Team Joined: Dec 2006 From: Lexington, MA Posts: 3,267 Thanks: 408  Here is an amusing little pattern . . . $\displaystyle \;\;\;\begin{array}{ccccc} 1 &=& 1 &=& 1^3 \\ 3 + 5 &=& 8 &=& 2^3 \\ 7+9+11 &=& 27 &=& 3^3 \\ 13+15+17+19 &=& 64 &=& 4^3 \\ 21+23+25+27+29 &=& 125 &=& 5^4 \\ \vdots && \vdots && \vdots \end{array}$ And a mathematical joke . . . $\displaystyle \;\;\;\begin{array}{ccc} 3^2 + 4^3 &=& 5^2 \\ 3^3 + 4^3 + 5^3 &=& 6^3 \\ \vdots && \vdots \end{array}$ 
April 2nd, 2014, 12:26 PM  #10 
Math Team Joined: Apr 2012 Posts: 1,579 Thanks: 22 
Nice!


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