Monday, February 8, 2016

Happy Chinese New Year 2016 (#LunarNewYear)


2016
= 12 × 168


     Wishing all my readers a Happy Chinese New Year ... or more accurately a Lunar New Year – many other Asians (e.g. the Japanese and the Koreans) also celebrate this festival.  This year  2016  is mathematically special, because it is the product of  12  and  168.  12  is a lucky number for Western people (e.g. 12 signs of the zodiac and 12 days of Christmas)  and “8” is especially auspicious in Chinese because it sounds like /  [fa in Mandarin,] which means to prosper or to grow.  168 sounds like 一路发 [yaad lou faat in Cantonese] which is to prosper all the way through life.

     I think this year will be challenging, because the Fire Monkey could be monkeying around even more with the economy and world peace.  Nevertheless if all human beings can unite together and learn to think critically, creatively and logically (and mathematics is about all these), the planet Earth can be a better place.  So best wishes to one and all!



Thursday, February 4, 2016

[EM_20160204PBIE] “Hillarious” Mathematics of Politics?

Problem
Six coins are tossed to decide a result for either “C” or “S”.  Assuming that the coins are fair, and that the results of the tosses are independent, calculate the probability that all the tosses are in favour of “C”.

Introduction
     Hot in recent news is the story of the purported six coin tosses that were needed to determine certain county delegates in the race between Hillary Clinton and Bernie Sandersin the state of Iowa.  All six coin tosses were in favour of Hillary Clinton, and the result is so improbable that some people said it was “hillarious”.
     How is the probability calculated?  This is an example of mathematics in real life events that has the potential to affect the United States of America, and the whole world (including Singapore).

Solution

Discussion
     In order to make the calculation, we make two assumptions: 
          (1) that the coins were fair, and
          (2) that the coin toss results are independent.
     So, what does it mean that the coins are “fair”?  It means that the probability of getting a “heads” is the same as the probability of getting a “tails”, which means ½ for each.
     Coin toss results can be “heads” or “tails”.  These are examples of events.  An event is something that can happen or not happen, and we associate a probability with it.  The probability is a number that indicates how likely the event happens.  It is between  0  and  1  inclusive.  Zero probability means a practically impossible event.  A probability of  1  means a practically certain event.  [The reason for me using the word “practically” is technical, which I shall not discuss.]  If the events do not affect one another (i.e. in our case, the coin tosses are not affected by the other coin tosses) then the events are said to be independent.  If the events are independent, then we can simply multiply the individual probabilites together, as above.  If the events are dependent, the calculation would be more complicated.

     So are the coin toss results valid?  I do not know.  All I can say is: improbable does not mean impossible.  Mathematics cannot tell whether the above assumptions (1) and (2) are correct.  But at least I “lay all the cards on the table”, so that astute students of probability know the basis of these calculations.  It is up to you to decide, but at least you would have made a mathematically-informed decision.  There could be other twists to the story, which is beyond the scope of this article.  This is one of the reasons why you need to learn mathematics carefully and think critically, whether or not you would become a mathematician,engineer, teacher or have a mathematics-intensive career.

H08. Make suppositions
H10. Simplify the problem
H11. Solve part of the problem

Suitable Levels
GCE ‘O’ Level “Elementary” Mathematics (Number patterns, with algebra)
* other syllabuses that involve probability
* anybody in the whole wide world!









Monday, February 1, 2016

[OlymLSec20160201PHHE] Pigeonhole Principle and Harry’s emails

Problem / Question

Handsome Harry has a secret email account that only four friends know.  Today he received 8 emails in that account. Which of the following is certainly true?
(A)  Harry received two emails from each friend.
(B)  Harry cannot have received eight emails from one of his friends.
(C)  Harry received at least one email from each friend.
(D)  Harry received at least two emails from one of his friends
(E)  Harry received at least two emails from 2 different friends.

Introduction
      This question is from some Kangaroo Mathematics Competition, which tests students on logic and not necessarily things from Singapore Mathematics syllabus. 

Solution
      (D)  Harry received at least two emails from one of his friends

Explanation
      This is an example of the Pigeonhole Principle.  Perhaps the easiest way to understand this is to imagine an array of pigeonholes with four columns (one for each of Harry’s friends) and pigeons (representing individual emails sent from the friends).  In the diagram below, I draw dots instead of pigeons.
As you can see, no matter how the eight dots / pigeons are placed, at least one of the friends will have at least two dots.  It is not possible for all the friends to have less than two emails.

Formal Proof
     We can use a proof by contradiction argument.  Suppose it were not true that Harry received at least two emails from one of his friends.  That would mean each of his  4  friends sent at most one email.  But then the total number of emails would be  4  or less.  This contradicts the given fact that Harry received  8  emails.  So this state of affairs is not possible.  Therefore, the opposite is true.  We conclude that Harry received at least two emails from one of his friends.

Final Remarks
      The Pigeonhole Principle is very useful in many situations, including computer science.  In general, if you have more objects (“pigeons”) than there are containers or slots (“pigeonholes”), one of the containers must have at least two of those objects.

H02. Use a diagram / model
H04. Look for pattern(s)
H05. Work backwards
H08. Make suppositions
H09. Restate the problem in another way

Suitable Levels
Lower Secondary Mathematics Competition / Olympiad
* other syllabuses that involve logic, combinatorics or Pigeonhole Principle
* any precocious or independent mathematics problem solver who is interested




Tuesday, January 26, 2016

[S1_20160126FMNT] Largest Common Remainder for Three Divisors

Problem / Question

Find the greatest 4 digit number that will divide 63, 45 and 69 so as to leave the same remainder.

Solution
     LCM + (smallest number – 1) = 7245 + 44 = 7289   J

Remarks
     This question is from National University of Singapore High School, which caters to students who are very interested in science and mathematics, and possibly an academic career.  Do not be fooled by my short and sweet solution.  The question is actually quite challenging, and I went by a long way before coming up with this elegant solution.  This reminds me of  Human Resource managers who think that more lines of code written by programmers means more work is done.  Actually, a lot of hard thinking could be involved in writing a one-line code that does the same job.
     Research shows that when expert problem solvers realise that they are stuck, they change tactics.  They go back to the drawing board.  “Insanity is doing the same thing over and over again and expecting the different results.” said Albert Einstein.  The five stages of mathematical problem solving are
     1. Understanding the problem
     2. Planning a strategy
     3. Executing the plan
     4. Evaluation
     5. Reflection (which can even include blogging about it!)
At the evaluation stage, if one finds that one is not getting the results, or if the approach is not elegant, one goes back to stage 2 to devise a new strategy.  I had tried using a complicated Chinese Remainder Theorem approach, got the solution after one page of work, and realised that the problem can be solved very simply.

How does the solution work
     The set of remainders dividing by  63,  45  and  69  repeat themselves in a cycle and the length of the cycle happens to be the Lowest Common Multiple (LCM).  We can see this quite easily: suppose  x  and  y  both give a remainder  R  dividing by  63,  45  and  69,  then  xy  gives a remainder of  0  when divided by  63,  45  and  69,  which means  xy  is a common multiple of  63,  45  and  69, of which the lowest is the LCM.
     It is a routine matter to get the LCM via prime factorisation as follows:   63 = 32 × 7,   45 = 32 × 5,  and   69 = 3 × 23.  We pick the highest power for each occurring prime and we obtain  LCM = 32 × 5 × 7 × 23 = 7245.
     The next thing to note is that remainders must be less than the divisors.  Hence the largest common remainder must be  44,  one less than the smallest divisor.  You cannot have a remainder of  45  when divided by  45,  because you could simply have bumped up the quotient (the result of division) by 1 and get zero remainder.  So from 7245, 7246, 7247, ... to 7289,  you get common remainders of  0, 1, 2, ..., 44.  Once you hit  7290,  the remainder for division by  45  will hit  0  while the remainders for  63  and  69  will be  45  but this will not be a common remainder.  The next time we get a common remainder will be  2×LCM = 2×7245 = 14490  which is  5 digits long.  So within  4  digits, the highest number with a common remainder is  7289,  which gives a common remainder of  44.

H04. Look for pattern(s)
H05. Work backwards
H09. Restate the problem in another way
H10. Simplify the problem
H13* Use Equation / write a Mathematical Sentence

Suitable Levels
Lower Secondary Mathematics (Sec 1 ~ grade 7) challenge
* other syllabuses that involve factors and multiples or number theory
* any precocious or independent learner who loves a challenge






Monday, January 25, 2016

[AM_20160125DAHT] Horizontal Tangents via Quadratic Discriminants

Problem

Introduction
     This is a Additional Mathematics textbook problem.  This question is of an intermediate level of difficulty.  The general method is by differentiation.  The equation of the curve happens to be capable of being put into a quadratic equation in  x.  Hence we can also use the theory of quadratic discriminants.  I present both methods of solution.

Method 1 (Using differential calculus)


Method 2 (Using quadratic discriminants)

Heuristics Used
H04. Look for pattern(s)
H05. Work backwards
H09. Restate the problem in another way
H11. Solve part of the problem
H12* Think of a related problem
H13* Use Equation / write a Mathematical Sentence

Suitable Levels
GCE ‘O’ Level Additional Mathematics
* other syllabuses that involve differentiation or quadratic discriminants
* any independent learner who is interested






Friday, January 22, 2016

[Maths_Education] The Need to Harness and Transcend Technology

Problem / Question

Help me please!  My teacher needs it tomorrow!
How many millilitres are there in 3.4 litres?

Answer
     LOL!  Well, if she really needs it, ask her to type the question into Google!

Remarks
     Actually, this is not a joke.  Many easy questions in school mathematics are now answerable by Google.  In fact, Wolfram Alpha is able to answer mathematically more difficult but routine questions.  Wikipedia and YouTube are also useful for learning mathematics.  There are many other good resources available on the Internet and public libraries.  The sad thing is,
1)
It seems that our current generation of kids views homework as a chore to be done for the teacher, not as an experience to be used for their own learning.  OK, maybe the homework task should have been designed better, to ask non-Googleable questions, but educators need to be aware of that is happening to our kids.
2)
Our kids do not know how to choose and use the abundantly available technology and resources
3)
Neither are they taught how to do this in school (do the teachers know it themselves?)
4)
This type of question merely targets the lower levels of Blooms Taxonomy
5)
Even for these easy questions, children are unable or unwilling to make the effort to find their own answers, and how are they to engage in Higher Order Thinking, creative thinking, reasoning etc.?  
     As technology evolves and improves and replaces many jobs, and as our children de-evolve and slacken, when this generation grows up they will not only face an even more challenging environment for their careers than today, they may not have the right values, attitudes and dispositions for living life.
     The purpose of learning mathematics in school should be to learn how to think and to serve others.  Human students must be educated to use technology and go beyond technology, to seek answers and to help other people instead of merely relying from other people for “help”.  This is one of the reasons why an identity (“learning to be a type of person”) approach to learning mathematics is important.

Tuesday, January 19, 2016

[S1_Expository] Exploring the HCF and LCM with a calculator


How does it work?
     When we reduce a fraction to its lowest terms, we actually cancel out as many common factors as possible.  So eventually the numerator and the denominator of original fraction get cancelled by their Highest Common Factor (HCF), a.k.a. “Greatest Common Divisor (GCD) ” in America.  Just as England and the United States are divided by a common language(*), we can divide  756  by  6  to get the HCF.  Why?  That is because  756  was divided by the HCF to get  6.  You may try a similar trick with the denominators.  You get the same conclusion.
     It is useful to know that the product of two numbers is equal to the product of their  HCF  and  LCM.  So   756 × 1386 = HCF × LCM.  Hence we have
Observe that the bracketed number  (756/126)  is equal to  6  which is the numerator of the reduced fraction.  So we do not even need to make that calculation.  Just take the reduced numerator  6  and multiply that with the original denominator  1386  to get the LCM.  You may try a similar trick with  1386  and  126.  You end up multiplying  756  by  11,  which gives the same answer.

Further Exploration
     Try the above with different pairs of numbers.  How can you extend this to find the HCF and LCM of three or more numbers?

(*) OK, just kidding.  In Singapore, we sort of follow British English, but we are flexible.

Suitable Levels
Lower Secondary Mathematics (Sec 1 ~ grade 7)
* other syllabuses that involve factors, HCF (GCD) and LCM
* any interested learner