# Was There a Pi Day on 3/14/1592?

In honor of Pi Day, here’s a bonus edition of Mean Green Math from two years ago.

March 14, 2015 has been labeled the Pi Day of the Century because of the way this day is abbreviated, at least in America: 3/14/15.

I was recently asked an interesting question: did any of our ancestors observe Pi Day about 400 years ago on 3/14/1592? The answer is, I highly doubt it.

My first thought was that $latex pi$ may not have been known to that many decimal places in 1592. However, a quick check on Wikipedia (see also here), as well as the book “$latex pi$ Unleashed,” verifies that my initial thought was wrong. In China, 7 places of accuracy were obtained by the 5th century. By the 14th century, $latex pi$ was known to 13 decimal places in India. In the 15th century, $latex pi$ was calculated to 16 decimal places in Persia.

It’s highly doubtful that the mathematicians in these ancient cultures actually talked to…

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# The Perfect Geometrical Christmas Present

Behold, a three-dimensional calendar on the faces of a rhombic dodecahedron:

# Faculty Office Hours

Kudos to Arizona State University for making this public service announcement.

# Pizza Hut Pi Day Challenge (Part 6)

On March 14, 2016, Pizza Hut held a online math competition in honor of Pi Day, offering three questions posed by Princeton mathematician John H. Conway. As luck would have it, years ago, I had actually heard of the first question before from a colleague who had heard it from Conway himself:

I’m thinking of a ten-digit integer whose digits are all distinct. It happens that the number formed by the first n of them is divisible by n for each n from 1 to 10. What is my number?

I really like this problem because it’s looks really tough but only requires knowledge of elementary-school arithmetic. So far in this series, I described why the solution must have one of the following four forms:

O , 4 O 2 , 5 8 O , 6 O 0,

O , 6 O 2 , 5 8 O , 4 O 0,

O , 2 O 6 , 5 4 O , 8 O 0.

O , 8 O 6 , 5 4 O , 2 O 0.

where O is one of 1, 3, 7, and 9. (The last digit is 0 and not an odd number.) There are 96 possible answers left.

Step 8. The number formed by the first eight digits must be a multiple of 8. By the divisibility rules, this means that the number formed by the sixth, seventh, and eighth digits must be a multiple of 8.

For the first form, that means that $8O6$ must be a multiple of 8. We can directly test this:

$816/8 = 102$: a multiple of 8.

$836/8 = 104.5$: not a multiple of 8.

$876/8 = 109.5$: not a multiple of 8.

$896/8 = 112$: a multiple of 8.

For the second form, that means that $8O4$ must be a multiple of 8. This is impossible. Let $O = 2n+1$. Then

$8O4 = 800 + 10 \times O + 4$

$= 800 + 10(2n+1) + 4$

$= 800 + 20n + 14$

$= 2(400 + 10n + 7)$.

We see that $7$ is odd, and therefore $400 + 10n + 7$ is not a multiple of 2. Therefore, $2(400 + 10n + 7)$ is not a multiple of 4 (let alone 8).

For the third form, that means that $4O8$ must be a multiple of 8. This is also impossible. Let $O = 2n+1$. Then

$4O8 = 400 + 10 \times O + 8$

$= 400 + 10(2n+1) + 8$

$= 400 + 20n + 18$

$= 2(200 + 10n + 9)$.

We see that $9$ is odd, and therefore $200 + 10n + 9$ is not a multiple of 2. Therefore, $2(200 + 10n + 9)$ is not a multiple of 4 (let alone 8).

For the fourth form, that means that $4O2$ must be a multiple of 8. We can directly test this:

$412/8 = 51.5$: not a multiple of 8.

$432/8 = 54$: a multiple of 8.

$472/8 = 59$: a multiple of 8.

$492/8 = 61.5$: not a multiple of 8.

In other words, we’re down to

O , 4 O 2 , 5 8 1 , 6 O 0,

O , 4 O 2 , 5 8 9 , 6 O 0,

O , 8 O 6 , 5 4 3 , 2 O 0,

O , 8 O 6 , 5 4 7 , 2 O 0.

For each of these, there are 3! = 6 ways of choosing the remaining odd digits. Since there are four forms, there are 4 x 6= 24 possible answers left.

In tomorrow’s post, I’ll cut this number down to 10.

# A request to the athletic department

Even though I’ve had nothing but good professional relationships with the athletic department at my own university, I still think this is really funny.

Source: http://imgur.com/vJWkqhe

# Shuffling cards

I really enjoyed this article concerning the mathematics of shuffling a deck of playing cards, justifying the claim that no two properly shuffled decks of cards have ever been the same: http://www.matthewweathers.com/year2006/shuffling_cards.htm

# 100,000 page views

I’m taking a one-day break from my usual posts on mathematics and mathematics education to note a symbolic milestone: meangreenmath.com has had more than 100,000 total page views since its inception in June 2013. Many thanks to the followers of this blog, and I hope that you’ll continue to find this blog to be a useful resource to you.

Twenty most viewed posts or series (written by me):

Twenty most viewed posts (guest presenters):

1. Engaging students: Classifying polygons
2. Engaging students: Congruence
3. Engaging students: Deriving the distance formula
4. Engaging students: Distinguishing between axioms, postulates, theorems, and corollaries
5. Engaging students: Distinguishing between inductive and deductive reasoning
6. Engaging students: Factoring quadratic polynomials
7. Engaging students: Finding x- and y-intercepts
8. Engaging students: Laws of Exponents
9. Engaging students: Multiplying binomials
10. Engaging students: Order of operations
11. Engaging students: Pascal’s triangle
12. Engaging students: Right-triangle trigonometry
13. Engaging students: Solving linear systems of equations by either substitution or graphing
14. Engaging students: Solving linear systems of equations with matrices
15. Engaging students: Solving one-step and two-step inequalities
16. Engaging students: Solving quadratic equations
17. Engaging students: Square roots
18. Engaging students: Translation, rotation, and reflection of figures
19. Engaging students: Using right-triangle trigonometry
20. Engaging students: Volume and surface area of pyramids and cones

If I’m still here at that time, I’ll make a summary post like this again when this blog has over 200,000 page views.

# Danica McKellar from NOVA

I really enjoyed this:

# Why Reluctant Students Still Should Learn Math

I love this quote from Math With Bad Drawings about why students should learn math:

In every walk of life, humans need to reason. So of course, they can learn these intellectual skills in other places. You don’t need math. But gosh, does math make it easier!

You can learn to taxonomize in biology, by considering the classification of organisms. But your taxonomies will never be perfect, because life doesn’t fit into neat little boxes. (I’m looking at you, protists.)

Life doesn’t… but math does.

Or you can learn to dissect arguments in civics. But emotions will flare. It’ll be tough to agree on premises. And even if you do, words like “justice,” “freedom,” and “common good” are subject to fuzzy interpretations and subtle misunderstandings. All words are like that: a little vague, tricky to pin down.

Except in math.

Logic shows up everywhere. But in math, it’s the whole game. Math isolates the operations of logic and reason so that we can master them.

In short: math is the playground of reason.

# Computer Cracks 200 Terabyte Math Proof

Here’s a cute problem, called the Boolean Pythagorean Theorem problem. Here are the first few Pythagorean triples:

$3^2 + 4^2 = 5^2$

$6^2 + 8^2 = 10^2$

$5^2 + 12^2 = 13^2$

$9^2 + 12^2 = 15^2$

$8^2 + 15^2 = 17^2$

$12^2 + 16^2 = 20^2$

$7^2 + 24^2 = 25^2$

$15^2 + 20^2 = 25^2$

$10^2 + 24^2 = 26^2$

$20^2 + 21^2 = 29^2$

$18^2 + 24^2 = 30^2$

$16^2 + 30^2 = 34^2$

$21^2 + 28^2 = 35^2$

$12^2 + 35^2 = 37^2$

$15^2 + 36^2 = 39^2$

$27^2 + 36^2 = 45^2$

$9^2 + 40^2 = 41^2$

$27^2 + 36^2 = 45^2$

OK, let’s have some fun with this. Let’s write every multiple of 5 (5, 10, 15, 20, 25, 30, 35, 40, 45) in boldface:

$3^2 + 4^2 = {\bf 5}^2$

$6^2 + 8^2 = {\bf 10}^2$

${\bf 5}^2 + 12^2 = 13^2$

$9^2 + 12^2 = {\bf 15}^2$

$8^2 + {\bf 15}^2 = 17^2$

$12^2 + 16^2 = {\bf 20}^2$

$7^2 + 24^2 = {\bf 25}^2$

${\bf 15}^2 + {\bf 20}^2 = {\bf 25}^2$

${\bf 10}^2 + 24^2 = 26^2$

${\bf 20}^2 + 21^2 = 29^2$

$18^2 + 24^2 = {\bf 30}^2$

$16^2 + {\bf 30}^2 = 34^2$

$21^2 + 28^2 = {\bf 35}^2$

$12^2 + {\bf 35}^2 = 37^2$

${\bf 15}^2 + 36^2 = 39^2$

$27^2 + 36^2 = {\bf 45}^2$

$9^2 + {\bf 40}^2 = 41^2$

$27^2 + 36^2 = {\bf 45}^2$

For nearly all of these equations, there is one number that’s in boldface and one that’s not. However, there’s one that is all in one typeface: ${\bf 15}^2 + {\bf 20}^2 = {\bf 25}^2$.

So here’s a question: is it possible to divide the integers so that every Pythagorean triple (not just the small ones listed above) has at least one number in boldface and another that’s not?

This May, it was proved that it’s impossible. The proof is very brute-force (from https://cosmosmagazine.com/mathematics/computer-cracks-200-terabyte-maths-proof):

The team found all triples could be multi-coloured in integers up to 7,824. As soon as they hit 7,825, it became impossible.

But to prove a solution doesn’t exist, you need to try all possibilities. There are more than $10^{2300}$ ways to colour all those integers, so the scientists used a few mathematical tricks to reduce the number of combinations to trial to just under one trillion.

Two days later, with 800 processors at the University of Texas Stampede supercomputer crunching all possibilities in parallel, the team had their answer – no.

There is no way to colour the integers 1 to 7,825 in a way that leaves all Pythagorean triples multi-coloured, the team reported in arXiv.

I had to read this news article a couple of times to appreciate this: a supercomputer ran for two days on a supercomputer (without parallelization, computation time was 51,000 hours), producing an output file of 200 terabytes, comparable “to the size of the entire digitized text held by the US Library of Congress.” Wow.