PI Day, More Videos on Counting, "Odds and Evens"

Since pi day fell on a Sunday this year, we should still be celebrating it today. Besides, March should be declared pi-Month!


It is always fascinating to see how readership (or should I say one-time viewership) always picks up around March 14th every year! I feel obligated to add another pi Day activity or exploration in addition to those I've posted the past three years. By the way, the pi Day Scavenger Hunt is the most popular post by far and I'm not even the one who thought of that idea!


Despite the title of this post, I did not upload a video for this activity. However, there is another video on the MathNotationsVids Channel on YouTube.

Here is an investigation/exploration/activity for middle and secondary:

Part (A)
(i) List all ordered pairs of positive integers (m,n) such that
(1) 1 ≤ m ≤ 10 and 1 ≤ n ≤ 10
(2) m and n are divisible by the same prime p

For example, (m,n) could be (6,9) since 6 and 9 are each divisible by the prime 3.

(ii) Should (9,6) also be counted?

(iii) Another way of expressing Condition (2) is:
The _______________ of m and n is ________  one.
Answer: gcf; not equal to or greater than

(iv) If you listed and counted correctly, you should have found there are 37 ordered pairs which satisfy both conditions. If not, have a partner check your list. Each of you should be checking each other's lists routinely.

Part (B)
(i) Explain, using the multiplication principle, why there are 100 ordered pairs which satisfy Condition (1) above.

(ii) ) What % of all the possible ordered pairs from Condition (1) are relatively prime. If you have immediate access to the internet, research this term before asking your teacher what it means!

(iii) In probability terms, you could say:

If one of the 100 ordered pairs (m,n)  from Part (A) is selected at random, the probability that
m and n are relatively prime is ____%.

Part (C) (more advanced)

If you have access to a graphing calculator, such as the TI-84 or TI-Inspire, enter the following program into memory (call it RELPRIME):

:ClrHome
:Prompt N
:0 → K
:For (X,1,N)
:For (Y,1,N)
:If gcd(X,Y) ≠ 1
:K+1 → K
:End
:End
:Disp K
:Stop

Using this program, complete the following table:

N..........Total # ord. prs..........# of not rel prime prs........% rel prime prs

10.........100.............................37....................................63%

20.........400............................ 145.................................

30

40

50

100

Notes:
K represents the count of ordered pairs which are not relatively prime
N represents the greatest value for the integers
gcd is found by going to MATH, then NUM, then 9:gcd(
The program slows down considerably as N increases. For N = 10, it checks 100 ordered pairs which may take only 2-3 seconds. For N = 100, it checks 100^2 pairs, which could take up to 4-5 minutes. Be patient!!

Conclusion: So what does all of this have to do with π ?
Well, as N increases without bound in the program, the probability that a randomly chosen ordered pair of positive integers (with values up to an including N) will be relatively prime approaches 60.7% rounded.

From out of the blue, compute 6/π²...
Want to know why? Well, that requires some advanced machinery involving infinite products, infinite series, and the Riemann Zeta Function! Perhaps, I'll do an informal development in a video. I love this stuff...


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"All Truth passes through Three Stages: First, it is Ridiculed...
Second, it is Violently Opposed...
Third, it is Accepted as being Self-Evident."
- Arthur Schopenhauer (1778-1860)


You've got to be taught
To hate and fear,
You've got to be taught
From year to year,
It's got to be drummed
In your dear little ear
You've got to be carefully taught.
--from South Pacific