[S1_20160227FZCK] Factorisation by Chunking

Problem / Question

 

Solution

Commentary

     Here I illustrate the usefulness of chunking to factorise (AmE: factor) an algebraic expression.  Observe that  3a – 2b  is a repeated part of the expression.  I call it a “chunk”.  To make it clear, I rewrite  (3a – 2b)²  as   (3a – 2b)(3a – 2b)  so that you can see it as two copies of the same chunk.  I highlight in yellow one copy of  (3a – 2b)  from each of 
(3a – 2b) (3a – 2b)   and  -3(3a – 2b).  The remaining stuff are highlighted in blue and green.  Take out the yellow chunk as common factor by writing it out on the left in the third line, shown in yellow.  You can pull out the common factor by writing it out to the right if you want, but here I chose to put it on the left.  The result would be equivalent anyway.  Once you have written out the common factor,  you write out the other stuff (shown highlighted in blue and green) into another other bracket.

     Once you understand how it works, you can actually do the second line mentally and write down the answer straightaway.  Chunking is a very useful technique in mathematics.  Here are some more examples of the technique of chunking: (1), (2), (3).

Heuristics Used

H04. Look for pattern(s)

H10. Simplify the problem

H11. Solve part of the problem

Suitable Levels

* Lower Secondary Mathematics (Sec 1 ~ grade 7)

* GCE ‘O’ Level “Elementary” Mathematics

* other syllabuses that involve algebra and factorisation (factoring)

* any learner who is interested

[Pri20160223FPDM] Pernicious Portion Problem? Shift Happens!

Problem / Question

Strategy

     This seems to be a confounding question on decimals.  What shall we do with the triangles?  Is there a short cut?

     Yes!  What you can do is to imagine putting the two triangles together to form a rectangle.  And then the solution becomes easy!  This is because the area is unchanged and hence the proportion of the shaded area is unchanged, is the same as before.  We can make use of fractions and convert it to a decimal.

Solution

Heuristics Used

H02. Use a diagram / model

H04. Look for pattern(s)

H09. Restate the problem in another way

H10. Simplify the problem

Suitable Levels

* Primary School Mathematics

* other syllabuses that involve fractions and decimals

* any learner who is interested

[Maths Education] Mathematical Journalling

     Nowadays, I see some schools / textbooks asking students to search the internet and to write on a certain problem on their “mathematics journal”.  It's so very guided.  It's so artificial.  The questions should come from the learners themselves, out of their own curiosity.  The learn then seeks to answer their own questions.  The journal can serve as to document and summarise their process of learning.

     The best maths journals are self-initiated.  Great mathematician Karl Friedrich Gauss and renowned scientist Richard Feynman kept math journals on their own accord, not because some teacher told them to do it. 

     When I was a student, I borrowed books from the National Library on things out of the normal curriculum.  I kept notes of things I learned.  I also did my own investigations.  I accidently discovered quadratic equations when I was in Primary 4.  I read guidebooks, asked my friend's elder brothers and sisters, my Chinese teacher (!) and other people to find out more.  I did not like factorisation by trial-and-error.  Neither did I like completing the square nor using the quadratic formula.  So I did my own research to find a sure-fire way to factorise without trial-and-error.  I finally managed to find a way, but my method had an uncanny similarity to the quadratic formula.  It was a Pyrrhic victory, but it was fun!  I thoroughly enjoyed it.

     If students need to be told or goaded to write mathematics journals, then we as educators need to ask ourselves:  Why?  What is their conception of mathematics and education?  What experiences have they gone through that lead them to these beliefs?

     Some food for thought, eh?

[P6_20160217RTTU] Books on Bookshelves

Problem

Introduction

     Here we have a numerically challenging problem that involves ratios, and it ultimately reduces to an algebraic problem with two unknowns.  Nevertheless, we are spoilt for choice as regards to methods of solution:-

     (1)   Bar Diagram Modelling

     (2)   explicit letter-symbolic Algebra

     (3)   “p” and “u”  (parts and units)

     (4)   Distinguished Ratio Units

     Despite the fact that Bar Diagram Modelling made “Singapore mathematics” famous, let us remember that it is only one of the ways of solving problem by diagramming, which is just one of the eleven Primary School heuristics recommended by the Singapore Ministry of Education.
     The methods have a lot in common, and they differ mainly in the form of presentation.  However, standard Bar modelling is impractical under high-stakes high-stress examination conditions for this problem, not least because one would have to cut the bars into many pieces.  One should not cut off one’s feet just so as to fit the shoes (削足适履), as one Chinese saying goes.  We need to be flexible and open-minded.  I present a solution using my own Distinguished Ratio Units.

Solution

Ans:  735 books

Commentary

     First off, we need to equalise the numerators of  ²/₅  and  1¹/₄ = ⁵/₄  and put them ratio form.   This is because the  “2”  in the  ²/₅  represents the same quantity as the  “5”  in  ⁵/₄.

We do this adjustment by multiplying the former through by  5  and the latter through by  2.  Thus we deduce that the original number of books in A and in B are  25  and  8  “heart” units respectively. 
     Next, we add on the  2  and  3  “triangle” units.  By doing a comparison, we can figure out that  1  “triangle” unit must be  45  more than  17  “heart” units.  So  2  “triangle” units must be equal to  34  “heart” units plus  90.  Replacing the  2  “triangle” units (shown in yellow) with their equivalent, we now know that  59  “heart” units plus 90 gives  444.  This allows us to figure out that  1  “heart” is actually  6.  Thus, we can work out what  1  “triangle” unit, and then what  5 “triangle” units are worth.

Final Remarks

     Due to the difficulty of the numbers, the solution presented above is about as streamlined as I can make it to be.  

     There is another variation that can be used – equalising the “triangle” units (akin to the technique of elimination in standard algebra).  What we do is we multiply the group with total  444  by  3  and to multiply the group with total  489  by  2.  This would give  6  triangle units on each side.  Then we can compare the “heart” units and continue from there.  This way of proceeding is not for those who fear 4-digit numbers.

     If there are nicer or more elegant ways to tackle this question, I would definitely love to hear from you.

Heuristics Used

H01. Act it out

H04. Look for pattern(s)

H05. Work backwards

H06. Use before-after concept

H09. Restate the problem in another way

H11. Solve part of the problem

Suitable Levels

* Primary School Mathematics

* other syllabuses that involve whole numbers and ratios

* any problem solver who loves a challenge

[S1_20160117AFNS] Much Ado About Nothing?

Problem

 

Introduction

     Assuming no typing errors, this is a tricky Secondary 2 question involving an equation with algebraic fractions.  How to solve it?  How to present the solution?

Solution

Remarks

[1]   Once the LHS expression has no meaning, it would not even make sense to continue.

[2]   This is a proof by contradiction type of argument.

It turns out that not all algebraic equations are soluble (or solvable).  This problem is one case in point.  The “unknown”  m  cannot be ⁵/₂ because that would make the expression undefined.  But if you substitute any other value, you always end up with nonsense like “15 = 0”.  So no matter what, there is no solution.  In other words, there is no value of  m  that you can substitute into the equation that makes it a true statement.

Heuristics Used

H05. Work backwards

H08. Make suppositions

Suitable Levels

* Lower Secondary Mathematics (Sec 2 ~ grade 8)

* GCE ‘O’ Level “Elementary” Mathematics

* other syllabuses that involve algebra

* any learner who is interested in algebra

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发 / 發  [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!

[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.

Heuristics Used

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!

[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.

Heuristics Used

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