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.