Figuring out Figurative Numbers--a mini unit

Last week, my Honors Problem Solving Seminar students wrote a collaborative guest blog about their unit on Figurate (or figurative) numbers here.

Here are the handouts. They are mostly from a book I wrote 16 years ago called Investigating Discrete Math. Unfortunately, I did not publish the book. Fortunately, I use it a lot as a resource. 

Most of the students had already seen The Locker Problem, which I blogged about here. This is the first example of a figurate number investigation, though they may not know it at the time.

I gave them the handout below and had them work in pairs, investigating the patterns in oblong and triangular numbers. Interestingly, they can get the oblong number explicit formula and the triangular number recursive formula quickly, but they have a hard time with the oblong recursive and the triangular number explicit. 

What I like about the worksheet is it leads right into the sums of  counting and even numbers, and that the nth triangular number represents the sum of the first n counting numbers, and that the nth oblong number represents the sum of the first n even numbers.

Many of my students needed a hint on finding the explicit formula for triangular numbers. I drew a diagonal through the middle of an oblong and soon they discovered triangular numbers were half the oblong numbers. So oblong number formula is n(n+1) and triangular is that divided by 2.

Since they remembered the locker problem, we reviewed square numbers and showed that the nth square number represents the sum of the first n odd numbers.


I also talked about the story of Gauss whose teacher was trying to keep him out of trouble as a very young boy.  She told him to add the numbers from 1 to 100, thinking it would take him quite a while to do so.  As the story goes, Gauss had the answer within moments.  How did he do it?  Problems 1 and 2 demonstrate two quick ways to do it:

1.         Shown below is the list of numbers he was told to add.

            Notice it you add together the first and last number, you get 101, as shown above.

            a.  Add together the second and the second to the last number.

            b.  If you continue adding (i.e., 3 and 98), you continue getting the same number of 101.  
                 How many pairs of numbers sum to 101?

            c.  Find the answer to the problem using the information above.

            d.  Generalize this for adding together the numbers from 1 to n.

2.  Another interesting way to find the sum of the numbers from 1 to 100 is to list the numbers from 1 to 100 and then underneath it, list the numbers from 100 to 1, as shown below.  Let both sums equal X.

          X =   1 + 2 + 3 + 4 +…+ 97+98+99+100

          X = 100+99+98+97+…+  4  +3 + 2 + 1   

Now add each of the pairs of numbers that are underneath each other; for example, add together the 1 and 100, the 2 and 99, etc.  How many pairs of the same number do you get?  Note that this answer equals 2X.  Why?  Use this new method to solve for X, the sum of the numbers from 1 to 100.

Once we had all of that "figured" out, I gave them the Handshake Problem. Many knew it, but could not remember the formula. I try to encourage students not to remember the formula, but to get it from scratch.

It was pretty cool when they eventually realized that the sequence for the number of handshakes with n people: 0,1,3,6,10,15,... (i.e., for 1 person there are 0 handshakes, for 2 people there is 1 handshake, etc) is one off of the triangular numbers: 1,3,6,10,15, .... Thus, they developed the formula n/2(n-1) by substituting n - 1 into n for the triangular numbers formula. Or, they came up with the formula by realizing that for n people, each person shakes hands with n-1 people, but you need to divide that by 2 since when I shake your hand and you shake my hand, that counts only once not twice.

Next, I gave them the Pentagonal Numbers Investigation.



This was much harder, and again, I encourage you to look at the blog the students wrote here to see their thinking. But one student figured out that pentagonal numbers (by looking at the 3rd figure) is just the sum of the two nth triangular numbers + one (n-1)st triangular number - 1. Pretty cool!!

Finally, I gave them the Pizza Problem.


Students discovered the sequence to be 2,4,7,11,... which most recognized as the triangular numbers + 1, which was pretty cool. I did not do the Pyramid Numbers Investigation, but next year I will, as it is a nice intro to the Method of Finite Differences, which demonstrates how to solve explicit polynomial formulas without looking at patterns.

Here is the Pyramid Number Investigation:



And the Method of Finite Differences plus homework handout. Keep in mind that this was made 16 years ago and so the typeset is a little "old!"

Students really enjoyed that this was out of the normal "math" that they see in their other math classes. We also went to https://www.desmos.com/ and did some regression there as well. There are many ways to solve for the formulas!!

The highlight of this was twofold. One was the collaborative blog they wrote. But the second was when they were the teachers for the Algebra 1 classes, and taught them the classic handshake problem with a twist: The Fist Bump Problem, a la Sam Shah. It was amazing. I will blog about that soon.

~Lisa

My Problem Solving Students Guest Blog: A Unit on Investigating Figurative Numbers

Our math teacher created a new course elective at Saint Andrew’s School called Honors Problem Solving Seminar. The class is interesting and has taught us everything from grit to fun little math tricks. So far we have discussed visual and figurative numbers, which are numbers that can be represented by a regular geometrical arrangement or sequence of evenly spaced points. They are most commonly expressed as regular polygons, for example, triangles, squares, pentagons, hexagons, etc. and also known as polygonal numbers. The class was a nice addition to the schedule providing us with a very relaxed way to enjoy math. This is what we have learned in our first big unit about figurative numbers.

(Note: this blog was collaboratively written by the entire class, originally in a Google Doc. All handouts are posted in my blog here.)

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We first started with oblong numbers.  Oblong numbers are the number of dots that can make up a rectangle with its length one more than its width.

To get to an explicit formula that can represent the development of the patterns of the number of dots on each rectangle, we first started by observing the rectangles. The first one consists of two dots; two for its length and one for width. The second has three for length and two for width. The third has four for length and three for width. We found out that as the pattern keeps going on, the new rectangle has one more dot on both its length and width than its previous term. So first, we constructed a recursive formula, using the number of dots on the previous term to define the number of dots in the current term. Since the length and width are both added by one each time, we thought that the current term is the resulting number of rows would be n and the resulting number of columns would be n+1.

Thus, the explicit formula for oblong number is n(n+1)

Then we were given triangular numbers. Triangular numbers are the dots grouped together that make up an equilateral triangle. Initially, we were trying to come up a formula based on the pattern and number of dots in each term: 1, 3, 6, 10... Obviously, the pattern does not appear directly based on the number of dots in each consecutive term. However, with Mrs. Winer’s small hint of drawing a diagonal, we found out that each term of the triangular number is exactly half of the oblong number. So the formula is basically the formula for oblong number divided by 2.

The explicit formula for triangular number is n(n+1)/2

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Now we can use what we have learnt from the oblong and triangular numbers to apply it to finding the sum of n counting numbers. If we pay attention to the triangular number, we will notice that the formation for triangular number is:

So we can find out nth triangular numbers means the sum of the nth counting numbers.

Now, for the sum of nth even numbers, (not counting zero), the first even number, 2 is two times bigger than one, the second even number, 4 is two times bigger than two, and the third 6 is two times bigger than 3.The pattern is the same for their sums. While the sum for all the counting numbers are triangular numbers, by studying the even numbers, we learned, that oblong, the combination of two triangular numbers, is the sum for the even numbers. We also can find it out by investigating the oblong numbers diagram.

After learning about the sum of n even counting numbers, we asked what the sum was for n counting odd numbers. We found out that you could find the sum of the odds just as fast as the sum of the evens. By looking at the diagram below (square numbers), we can conclude that the explicit equation for the sum of odd numbers is  n2 because in the diagram the number of dots in each picture is a perfect square of the nth term.

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To mix it up a bit, our math teacher presented to us the Pizza Problem.

Watch the video below for the Pizza Problem, plus an explanation of the Method of Finite Differences.

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Moving right along, we now faced the challenge of pentagonal numbers, which served to extend and solidify the concept of triangular numbers.

           

In a pentagon diagram, we can divide it into three parts (shown in the picture): red, blue and green. Count the dots in each part, and we can see the red part and the green part are the same; they are both the nth triangular number. In the middle, the number of dots in the blue section is also triangular number, but instead of nth term, it is the (n-2nd) term. Then we add all three parts together, but we count the first dot at the bottom twice, so we subtract 1. Simplify the equation, and we get the pattern for the pentagonal numbers.

RED:

1     1       1

2     3       1+2

3     6       1+2+3               -->    

4     10    1+2+3+4

5     15    1+2+3+4+5

…     …     …

GREEN:

1 1          1

2 3          1+2

3 6          1+2+3              -->

4 10       1+2+3+4

5 15       1+2+3+4+5

… …        …

BLUE:

1 0          0

2 0          0

3 1          1                    -->

4 3          1+2

5 6          1+2+3

…    …         …

Final Simplification

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Seeing that we were able to complete each task given so far, Mrs. Winer then challenged the class with yet another problem that once again would require an understanding of both the oblong and triangular numbers. She asked us,

“ What’s the difference between the sum of the first 2015 even counting numbers and the sum of the first 2015 odd counting numbers.”

At first the class was very confused and did not know how to solve this problem, but after much hard work, our class proudly could say that we discovered not one but TWO! very different methods to solve the problem.

Solution 1

First line up the first 5 even and first 5 odd numbers

2 + 4 + 6 + 8 + 10

1 + 3 + 5 + 7 + 9

Now you can see that the even numbers are always one greater than the odds. Since there are 2015 numbers, and each time the evens are one number more than the odds, the total difference will be the 2015.

Solution 2

This method uses direct substitution to find the first n counting even and odd numbers.

“N^2+N,” is the equation for sum of the even numbers and “N^2” for the sum of the odd numbers. The difference between the first n even and n odd counting numbers is the difference between the nth oblong number N^2+N and N^2:

N^2+N - N^2

The class then used direct substitution to solve the problem

(2015^2 + 2015) - (2015^2)= 2015

Although the two methods are very different, they both work in finding the answer. The class was split over which one they thought was easier.

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According to Pythagoras,“everything is related to mathematics. Numbers are the ultimate reality, and through mathematics, everything can be predicted and measured in rhythmic patterns or cycles.” So far, Pythagoras has been right about everything we have learned in this class and it is cool to see how everything comes together. Although there may be different ways to solve a problem, when it comes to math there is no ambiguity and everything eventually will fit together to form one solution.