# IntMath Newsletter: Valentine’s, applets, maps

By Murray Bourne, 14 Feb 2019

14 Feb 2019

In this Newsletter:

0. Geometry of love

1. New on IntMath: 3 new interactives

2. Quirky measurements

3. Resources: Map projections

4. Math movies: Speed of light, puzzles

5. Math puzzle: Remainders

6. Final thought: Finding your soul mate

## 0. Geometry of love

Happy Valentine's, everyone! Here's a mathematical interpretation of the ubiquitous symbol of love.

Did you ever consider the geometry of the idealised heart shape used in countless Valentine's messages? It's just a square rotated 45°, connected to 2 semicircles.

## 1. New on IntMath

### (a) Math Art in Code: Animated Moore Curve

This article from the journal *Nature* includes an interesting application of space-filling curves to energy storage devices:

Bioinspired fractal electrodes for solar energy storages

### (b) Degree and roots of polynomial equations

The concepts behind polynomial equations are quite important in mathematics, but a lot of students get bogged down in the algebra, while not seeing the big picture

I rewrote the opening section and have included an interactive graph, that helps explain the concepts. |

The rest of that chapter introduces the Remainder and Factor Theorems, which I've always felt were rather useless, since only a very limited subset of polynomial equations can be solved using those techniques (the degree has to be low, and the numbers have to be "nice" so factoring is easy. But what about the huge number of cases where the numbers don't lead to easy factoring by inspection (or trial and error)?

Of course we should sove such cases using computers!

### (c) Euler's Identity exploration

Leonhard Euler was a brilliant Swiss mathematician who made many contributions to the fileds of physics, astronomy and engineering.

His famous identity, *e*^{iπ} + 1 = 0, has captivated people (since it neatly involves seven fundamental mathematical symbols) and proved to be very useful for simplifying many mathematical processses.

This page allows you to explore the concepts behind the identity |

THis page was a request from reader Owen.

## 2. Quirky measurements

Each of these seemingly informal measurements actually have a scientific component to them. |

## 3. Resources - map projections

### (a) Map Projections on the Web

Every 2D map of the world has some distortion, the inevitable result of trying to project a 3-D object onto a 2-D plane.

Some map types are better than others for preserving area, or shape, but none are prefect.

GeoViz Studio has some great tools for exploring different projections. See especially the third (final) interactive example where you can choose from around 100 different projections. Go to: Map Projections on the Web |

GeoViz Studio also provides some great interfaces for creating your own map projections. See: Get Started Creating D3 Maps

These could be good starting places for students who want to learn how to code, via interesting challenges.

### (b) Dymaxion Buckminster Fuller map

Inventor, architect, designer and futurist Buckminster Fuller proposed the Dymaxion projection in 1943.

Many of his designs were based on the geodesic sphere, an imaginary tesellation of a sphere into triangular pieces.

His Dymaxion Map is a projection based on the above sphere (actually basesd on an icosahedron).

See also the results of a 2013 competition to re-invent Fuller's map:

7 Brilliant Reinventions of Buckminster Fuller's Dymaxion Map

## 4. Math Movies

### (a) Speed of light

These 3 simple NASA animations show how slow the speed of light really is, when we consider astronomical scales. This article is from Business Insider: |

### (b) The joyful, perplexing world of puzzle hunts

This video reminds us the human brain is wired to solve problems, and we experience a rush when we are successful, especially if it was challenging. |

One takeaway is we don't give students enough opportunity to **create the problems** in math class - they learn a lot by doing so!

## 5. Math puzzles

The puzzle in the last IntMath Newsletter asked "Who dies" in a scenario involving a rolling stone. Several conflicting answers were submitted. I added my own interpretation, which concludes D dies, but C may just have had a lucky day. Further comments are welcome!

### New math puzzle: Remainders

Find the least two positive integers having the remainders 2, 3, 2 when divided by 3, 5, 7 respectively.

You can leave your response here.

## 6. Final thought - using math to find your soul mate?

Fewer people are meeting the love of their life in a "natural" way and are turning to apps to make connections for them. Algorithms continue to take over our lives and we all need to have a better understanding of the math behind such algorithms.

From Business Insider:

Here's how you can use math to find your soul mate — and why we're so resistant to that idea

Until next time, enjoy whatever you learn.

See the 17 Comments below.

14 Feb 2019 at 9:05 pm [Comment permalink]

Hi Murray,

Thanks again for lots of fun things to think about.

One quibble I have though is with your dismissal of the Remainder and Factor Theorems as "rather useless". On the contrary, I think they are the key to understanding how polynomials (and many other functions) "work" - even when the numbers are not "nice".

Just as one example, if you show me just one intercept of any cubic then I can factor the thing completely - and even if the intercept is not "nice" I can get a factorization of arbitrarily high accuracy on an arbitrarily large domain so long as you will let me zoom in as close as I like to read the approximate value of that intercept.

As another, if you can give me the exact value v of any polynomial, p, at any point, a, then I can reduce the problem of factoring p(x)-v to the the solution of a lower degree polynomial q where p(x)-v=(x-a)q(x).

All the best.

Alan

14 Feb 2019 at 10:34 pm [Comment permalink]

@Alan - good to hear from you, as always. My concern is with the impression given by most text books that you can solve any polynomial using the Remainder and Factor Theorems. However, you can't if the factors stray from the usual suspects: ±1, &pm2; ±3, and maybe ±0.5.

I totally agree the

conceptis important (whenever we can make a complex task easier by performing an interim step), but the mechanics of actually solving higher-order polynomials is always best left to the computer - either by direct solving, or by graphing and zooming in, as you mentioned.I just feel we put students through a process (if we follow the "text book way", which many teachers do) without being totally honest about the limitations of that process, and the better alternatives.

I agree with you - the concepts are important for knowing what an equation is, and what it's solution implies. But how we get that first intercept can take students ages - and is it really "good" for them?

(I once taught a numerical computing class that involved equation solving as one of the topics. In one tutorial, while most of the students were churning away at some tedious Remainder Theorem task I had given them, one student put up his had with the correct answer which he'd obtained very quickly by trial and error. This incident challenged my view.)

14 Feb 2019 at 11:02 pm [Comment permalink]

Murray Bourne - I forwarded your ('blog?) to many friends and former colleagues with this remark: Ann and friends -

What a fascinating and joyful website! And I just sort-of stumbled into it.

Robert

I would like to comment on the article re: Using Math to Find Your Soulmate. I wish I could be brief but it is not my nature.

I have been married twice. The first was happy enough, and produced two fine sons and five (so far) grandsons. However, there was always pushing and hauling over words like 'me, mine', et c.

My background is in physics, math and statistics, with most of my career spent in signal analysis (sonar, radar, cardiac medicine). In the '70s and '80s I ran a small research group doing cardiac signal analysis for disease detection and pacemaker control. We needed an additional programmer. Of the few who applied, one had a degree in statistics from UChicago and was clearly the outstanding applicant. Our relationship progressed from boss/employee to partner in sharing and developing ideas, to friends. The partnership and deep understanding was wonderful. I often asked myself why a marriage couldn't have a partnership like that.

Eventually, I left the company and took a job out of state. It soon became clear that my then-wife and family had no intention of joining me. During that first year, several friends from my former company came to visit me and see how I was doing - including my former partner. This was some eight years after we met. Eight years and one minute after we met, we fell in love. Up to that point we had never laid romantic eyes on each other.

My first wife and I agreed to divorce. I re-married about a year later to my partner and statistic major. We have now been married 30 years and counting. Our underlying theme is P.F.L. (Partners, Friends and Lovers), like Andrew Carnegie's three legged stool, the supporting structure for our entire relationship.

And all because of statistics.

14 Feb 2019 at 11:55 pm [Comment permalink]

I don't know how to do the problem in a clever way so I just used excel to look for numbers generated by which are the same.

Found 23 and 128 pretty quick by looking manually. If we had to find the first 100 I'd try some kind of matching formula

15 Feb 2019 at 2:44 am [Comment permalink]

Very interesting articles.Just loved each one of them. Many thanks.

15 Feb 2019 at 8:25 am [Comment permalink]

The least two integers having the remainders 2, 3, 2 when divided by 3, 5, 7 are, respectively:

(2, 5)

(3, 8)

(2, 9)

If the quotient is assumed to be zero when the dividend is less than the divisor, then the remainder satisfies

reminder (m/n) = m - n quotient (m/n)

15 Feb 2019 at 8:50 am [Comment permalink]

The 3 divisors are all prime numbers which means

the LCM will be the product of the primes = 105.

Sets which satisfy mods with given remainders of r.

mod 7 with r of 2 {2, 9, 16, 23, 30…}

mod 5 with r of 3 {3, 8, 13, 18, 23…}

mod 3 with r of 2 {2, 5, 8, 11, 14, 17, 20, 23…}

The least positive integer having remainders of

2, 3, 2 when divided by 3, 5, 7 respectively is 23

as shown above. Successive least positive integers

would be at 1k + 23, 2k +23, 3k + 23… where k = 105.

1k +23 = 105 + 23 = 128 which is the 2nd solution.

15 Feb 2019 at 11:32 am [Comment permalink]

23 and 128

16 Feb 2019 at 7:35 am [Comment permalink]

Answer: The two positive integers are 23 and 128.

16 Feb 2019 at 7:29 pm [Comment permalink]

Remainders Puzzle Question.

Investigation of the first few positive integers and their remainders shows that 23 is the smallest integer with the desired property. Observation of the cyclical nature of the remainders when integers 1-23 are divided by 3, 5 and 7, shows that the desired remainders will occur after 3x5x7 = 105 steps. Since 23+105=128, then 128 is the second positive integer.

The required integers are 23 and 128.

18 Feb 2019 at 3:14 pm [Comment permalink]

The numbers are 23 and 128.

Time to develop equation v trial and error ?

Opting for the latter, with the assistance of some observation.

1. For mod 5 to yield 3, number must end with 3 or 8.

2. For mod 3 and 7 BOTH to yield 2, mod 21 of the number must also yield 2.

Let the trial begin:

21 + 2 = 23 Pass = 1st Number

23 + 21 = 44 Last digit not 3 or 8

44 + 21 = 65 ditto

65 + 21 = 86. ditto

86 + 21 = 107 ditto

107 + 21 = 128. Pass = 2nd Number

Success in only 6 Tries. Lucky this time!

18 Feb 2019 at 9:21 pm [Comment permalink]

the two numbers are 233;443

18 Feb 2019 at 11:41 pm [Comment permalink]

23 and 128

19 Feb 2019 at 2:10 pm [Comment permalink]

finding a soul mate with math is a really fun idea. thank you for sending me your newsletter. I am looking forward to a lot of good times.

23 Feb 2019 at 5:07 am [Comment permalink]

Answer) 23 and 128

6 Mar 2019 at 6:35 am [Comment permalink]

Let Us assume that no be N such that

N=3q+2

N=5r+3

N=7p+2

Solving first two equations we get

3q-5r=1

Solving last two equations we get

7p-5r=1

Solving first and last question we get

7p=3q

Therefore by comparing

3q=7p=5r+1

Taking q= 7, p= 3 and r=4

N=23

Again Taking q= 42, p= 18 and r=25

N=128

Therefore least two numbers are 23 and 128

8 Mar 2019 at 1:22 am [Comment permalink]

I used a fairly straightforward excel file to solve this one in a fairly brute-forceish kind of way: a combination of the =Mod() function to give me the remainders, and an =If(And()) function to determine whether or not each of the three conditions had been met for each positive integer.

Once you’ve put together such a spreadsheet, you can very easily find all the positive integers that fulfil the requirements, starting with 23 and going on to 128, 233, 338 and so on.

But what’s going on mathematically? I must confess that I haven’t actually arrived at a totally satisfactory mathematical proof, but I do have a few insights as to what’s going on:

Our first and third divisors, i.e. 3 and 7, had to have the same remainder 2. If we multiply these divisors together we get 21, and then of course we need to add 2 on to get it as a remainder. It just so happens that 23 also gives the remainder 3 when divided by 5, which means that 23 is our first answer.

Our divisors are all prime numbers, which I do not believe is coincidental :-). If we multiply them all together, we get 105. Adding 105 to 23 gives us 128, which is the next number that satisfies our requirements. Thereafter you just keep on adding 105 to get as many integers as you like which also tick all our boxes. If our divisors weren’t all prime, there’s a very good chance that the pattern would repeat more frequently.

(Interestingly, the pattern also works if we take 105 away from 23 to give -82. Then -187, -292 and so on, so there are as many negative numbers that work as there are positive ones.)