[Pri1_20161021DIV] Labelling as a strategy for Division

Introduction

          The Singapore mathematics syllabuses are very well designed, especially the primary school syllabus.  Fundamental concepts and skills are introduced before going on to complex calculations and problem solving.  At primary 1, pupils learn the idea of multiplication and division of small numbers by grouping (or partitioning).  They are not made to recite the times tables meaninglessly.

          Division is easy if the number of things in each group is known.  You just keep on circling the known number of objects until everything is circled.  However, if the number of groups is required but the number of things in each group is not given, and if the objects are not arranged in a convenient way, the task can be a bit more challenging.  Remember: they have not memorised the multiplication tables yet.

Problem / Question

Solution (Suggested)

          One way to solve this problem is to label the fish 1, 2, 3, 1, 2, 3, ... in a cyclic fashion, assigning fish to each of the three friends one at a time, thereby ensuring that each person gets the same number of fish.  Start with “1” somewhere on the left, “3” on the right and “2” somewhere in the middle.  Assign the next “1” close to the previous “1”, the next “2” close to the previous “2” and the next “3” close to the previous “3”.  So all the 1s are close together, the 2s are close together and the 3s are close together.  After all the fish have been labelled, the partitioning (or grouping) becomes obvious.

Heuristics Used

H02. Use a diagram / model

H04. Look for pattern(s)

H09. Restate the problem in another way

Suitable Levels

* Primary School / Elementary School Mathematics

* any precocious or independent learner who is interested

[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  x – y  gives a remainder of  0  when divided by  63,  45  and  69,  which means  x – y  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.

Heuristics Used

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

[AM_20151226EIQR] Looking for a Pea among Quadratic Roots?

Question

Introduction

     This question is about finding the parameter  p, and not about solving for the “unknown”  x.  It is heavy on algebra, one has to be patient, careful and meticulous.  Please refer to this article for a recapitulation of (Vieta’s) theory of Quadratic Roots.

Solution

Heuristics Used

H04. Look for pattern(s)

H05. Work backwards

H13* Use Equation / write a Mathematical Sentence

Suitable Levels

* GCE ‘O’ Level Additional Mathematics

* other syllabuses that involve quadratic roots
* any learner who is interested

[AM_20151224QERI] Quadratic Roots and Use of Identities

Problem

Introduction

     Here is a fairly standard question on roots of quadratic equations, except that part (iii) is slightly more challenging.  To solve this question, one must know the square of sum identity well.

Recapitulation

     Please refer to this previous article  and  this article  for the theory on quadratic roots.

Solution

Heuristics Used

H04. Look for pattern(s)

H05. Work backwards

H09. Restate the problem in another way

H11. Solve part of the problem

H13* Use Equation / write a Mathematical Sentence

Suitable Levels

* GCE ‘A’ Levels H2 Mathematics

* other syllabuses that involve roots of quadratic equations

* any learner who is willing to learn

[Pri_20151223WNSV] Unravelling Four Whole Numbers

Problem

If  A,  B,  C  and  D  are whole numbers such that  A × B = 8,  B × C = 28, C × D = 63,  B × D = 36,  find the values of    A,  B,  C  and  D.

Introduction

     This question seems to be taken from a secondary school textbook from a chapter on linear equations.  However, I think a good  primary school pupil could attempt this.

Strategy

     The key to solving the above problem is to make observations.  When you multiply up the first two equations, you get an  A,  a  C  and two  Bs  in the product.  Hmmm ... This doesn’t look promising ...  Ah!  But when you multiply the second and the third equations together, you get an  B,  a  D  and two  Cs  in the product.  This can cancel (via division) with the fourth equation which has one B  and one  D  in the product.

Solution

Remark

     Always cancel as much as possible, to avoid large numbers and reduce chances of making careless mistakes.

     By the way, a whole number is a non-negative (zero or positive) integer that does not contain any fractional part.   As such, the set of whole numbers is {0, 1, 2, 3, 4, ...}.  Thus we do not need to consider the negative square roots.

Heuristics Used

H04. Look for pattern(s)

H05. Work backwards

H10. Simplify the problem

Suitable Levels

* Primary School Mathematics (Challenge)

* Lower Secondary School Mathematics (Challenge)

* other syllabuses that involve whole numbers

* anyone game for a challenge

[H2_VJC2015PromoQ10_IAXS] Volume of a Doughnut

Problem

Introduction

     This is a problem involving the calculation of the volume of solid of revolution of an enclosed region.   The junior colleges (or senior high schools) like to set this type of question. 

Technique

     If the axis of revolution is the  y-axis, the basic formula is   ò px² dy   with the appropriate lower and upper limits.  Notice that this only works for the region between one curve and the axis and when rotated, this will generate a solid with no hollow parts.  An enclosed region, however, consists of two curves.  In our case, when we make  x  the subject, we find that we have two choices.  One of them leads to a curve that is further away from the axis of rotation.  I call that the outer curve.  The other curve is the inner curve, and this is nearer the axis of rotation.  We need to subtract the volume generated by the inner curve from that generated by the outer curve.

Solution

Remarks

     In this example, the curve on the right happens to be the outer curve.  If the equation were

(x + 93)² + y² = 15²,  the circular region would be on the left of the  y-axis and the outer curve would be on the left.

     For your information, the above solid of revolution is a torus.  This is the shape of a doughnut, (or hoopla-hoop, circular tube, or Polo mint perhaps?).  It is the inner curve that gives the hole in the “doughnut”.

     You can imagine in your mind’s eye that as the circular disk revolves around the  y-axis, its centre traces out a circular path of  93 units.  By the Second Centroid Theorem of Pappus,

     volume = length of path of centroid × area of cross section = 2p(93) × p(15)² =  41850p²

In general, the volume of a torus with major radius  R  and minor radius  r  is

     volume = 2pR × pr² =  2Rr²p²

If you know this fact, you use it to check your calculations.  Although this is not in the H2 Syllabus, but it is something interesting to explore.

Heuristics Used

H02. Use a diagram / model

H04. Look for pattern(s)

H05. Work backwards                                [e.g. making  x  the subject]

H09. Restate the problem in another way  [symmetry: volume is twice of upper half]

H10. Simplify the problem                         [integration by substitution]

H11. Solve part of the problem

H13* Use Equation / write a Mathematical Sentence

Suitable Levels

* GCE ‘A’ Levels H2 Mathematics

* IB Mathematics HL (Applications of Integration)

* Advanced Placement (AP) Calculus AB & BC

* University / College Calculus

* other syllabuses that involve Applications of Integration

* any precocious or independent learner who loves to learn

[MathEd] Critique: Problematising Mathematics Education

In Response to Article
The Politics of Math Education


My Comments

1. It is good to problematise mathematics education.  Certainly politics is involved in the choice of mathematics curricula.

2. However, is it good to argue and debate so much that nothing gets done?  Are you chasing down false dichotomies?  Are you assuming that you cannot have it all?

3. In Singapore, we do not argue so much and students go on to perform well in "mathematics".  Unfortunely, I feel, they get a very narrow and distorted view of what mathematics really is.

4. What is the answer the dilemma?  I think it's a question of identity.  I think students should make well-informed negotiated decision about the kinds of people they want to be and how a wholistic mathematics education serves to develop them not just in terms of skills and content, but also in terms of values, habits / dispositions, problem solving ability and critical thinking, ... etc.  I would be interested to learn of and even work with curriculum planners heading in this direction.

[PriOlym_20151130RTAC] Ratio with One Circle Overlapping Two

Question

Introduction

     This question is like this previous one, except it is of olympiad standard.  I illustrate the solution of this without algebra, by using Distinguised Ratio Units.  As before, I try to match parts to an equal number.  But here we have quite a mixture of different types of units.

Solution

Commentary

     Basically we make the triangle units to number 12 and do the same for the circle and square units.  It turns out that one triangle unit is the sum of one circle unit and square unit.  We deduce that 9 circle units (for the area of A) plus 6 circle units (for the area of B) is the same as 8 circle units and 8 circle units.  The reduction of circle units must be equally compensated by the increase in the circle units.  Thus one circle unit is the same as two square units.  From here, things become easy.

Heuristics Used

H02. Use a diagram / model

H04. Look for pattern(s)

H05. Work backwards

H09. Restate the problem in another way

H11. Solve part of the problem

Suitable Levels

* Primary School Olympiad Mathematics

* Primary School Mathematics (challenge)

* other syllabuses that involve areas and ratios

* anyone who is game for a challenge

[EM20151129CGCB] Bisector of a Chord in a Circle

Problem

Introduction

     This “elementary” mathematics question poses a challenge because it actually testing Coordinate Geometry and Circle Geometry.  In the setting of tests and exams, there seems to be a trend of combining topics.  To solve this problem successfully, students need to know that when a chord is bisected (cut into two equal parts), the line segment joining its mid-point to the centre of the circle will be perpendicular to the chord itself.  Thereafter, we can proceed with Pythagoras’ Theorem.

Solution

Remarks

     There is actually no boundary between topics and even subjects.  Things to be learned are separated into topics only to facilitate teaching of the material.  Students are encourage to adopt a more wholistic view of knowledge.

Heuristics Used

H04. Look for pattern(s)

H05. Work backwards

H09. Restate the problem in another way

H10. Simplify the problem

H11. Solve part of the problem

H13* Use Equation / write a Mathematical Sentence

Suitable Levels

* GCE ‘O’ Level Mathematics (“Elementary Mathematics”)

* GCE ‘O’ Level Additional Mathematics (revision)

* other syllabuses that involve geometry and coordinate geometry / analytic geometry
* whoever is interested

[Pri20151129RTAO] Equalising Ratio Units for The Overlap

Question

 

Introduction

     This is a primary school ratio problem that is quite a favourite among question setters, but poses headaches for pupils and parents.  The trouble is that the ratios use different base units and this makes it difficult to compare the ratios.  Can we avoid using algebra or trial and error?  

Strategy

     Note [H04, H09] that the difference in the areas between the rectangle and the square (including the shaded overlapping part) is exactly the same as the difference between them without the overlapping part.  With this crucial observation, we can proceed to try to equalise the ratio units [H10] of the aforementioned differences.  This can be done by multiplying to get to the Lowest Common Multiple, which, in this example is 6.  Henceforth we can be sure of using the same ratio units, because the same number of units are used to refer to the same quantity.

Solution

Summary

     Ratio problems are solved by making sure that we use the same type of units.

Heuristics Used

H02. Use a diagram / model        [ table ]

H04. Look for pattern(s)

H05. Work backwards

H09. Restate the problem in another way

H10. Simplify the problem

H11. Solve part of the problem

Suitable Levels

* Primary School Mathematics

* other syllabuses that involve areas and ratios

* anyone who wants to learn

[AM_20151127DF2D] Differentiation with Chunking and Elimination

Question

Introduction

     Although this looks like a differential equation question, the student is not required to solve the differential equation.  The requirement is just to derive the equation.  This would be a challenging question for secondary 4 (~ grade 10) students taking Additional Mathematics or their counterparts in Integrated Programme schools.

Strategy

     One way to do this is to differentiate the given equation once and again and just verify the equation by substitution.  The problem is that when we repeatedly apply the Product rule

the terms tend to sprawl.  A way to keep things neat is to try to recognise chunks and also use elimination.

Solution

Remarks

     After differentiating once, we notice that  10xe^2x   is twice of  5xe^2x,  and this allows the simplification in [1].  The second differentiation yields  10e^2x   which, we notice, is twice of  5e^2x.  We can get rid of that term.   Multiplying equation [1] by 2 gives  10e^2x  in equation [3],  which matches nicely with the same term in  [2].  So we can eliminate that term via elimination.  After that, we just need to rearrange things to get the final equation.

Heuristics Used

H04. Look for pattern(s)

H05. Work backwards

H10. Simplify the problem

H11. Solve part of the problem

H13* Use Equation / write a Mathematical Sentence

Suitable Levels

* GCE ‘O’ Level Additional Mathematics, “Integrated Programme Mathematics”

* GCE ‘A’ Levels H2 Mathematics (revision)

* AP Calculus AB / BC (revision)

* University / College calculus (revision)

* other syllabuses that involve differentiation

* any learner interested in calculus

[H2_20151125DEST] Differential Equation via Substitution

Question

Introduction

     This question was taken from a Facebook group.  The suggested substitution was added in to make it accessible to students taking H2 Mathematics.  In the original question, no suggested substitution was given.

Strategy

     Observe that the given equation has  x  and  y.   After substitution, we should get an equation with  v  and  x only.  How to make  y  “disappear”?  One way is recognise patches that can be substituted for  v.  Another way is to make  y  the subject and differentiate that with respect to  x.  Replace the derivative ^dy/_dx  with your new expression.  Here, I do both at once!

Solution

Remarks

     Usually we try to express  y  in terms of  x,  but here, it is more convenient to express  x  in terms of  y.

     Since  A  is an arbitrary constant,  A+1  is still an arbitrary constant.  Also, whenever there is a “ln” appearing in the solution, it is good to introduce “ln” with an arbitrary constant.  So we can lump  A  and  1  together and call it  ln B.  Nothing is lost in this process.  ln B  is able to achieve all possible numbers.   ln B  is negative if  B  is a fraction between  0  and  1,  and is positive if  B  is more than  1  and is zero if  B = 1.

     This is not an applied differential equation, so there are no exogenic reasons or hints for us to suppose that  x + y + 2  is a positive quantity.  As such, we still need to leave it as | x + y + 2|.  To get rid of the absolute or modulus sign properly, we can replace  ±B  with  C.

Heuristics Used

H04. Look for pattern(s)

H05. Work backwards

H09. Restate the problem in another way

H10. Simplify the problem

H11. Solve part of the problem

H13. Use Equation / write a Mathematical Sentence

Heuristic that cannot be used here

H08. Make suppositions

Suitable Levels

*  GCE ‘A’ Level H2 Mathematics

*  IB HL Mathematics Calculus Option

*  Advanced Placement (AP) Calculus BC

*  university / college calculus

*  other syllabuses that involve differential equations

*  anyone who is game for a challenge

[S1_20151124AESR] Slanted Rectangle does not need Pythagoras

Question

Introduction

     This is another “Bonus Question” at a secondary level from somewhere that the question poser did not mention, but I guess it is most likely an Integrated Programme school in Singapore.  It is a beautifully crafted question.  The presence of a slant line seems to necessitate the usage of Pythagoras’ Theorem.  However, we have seen that Pythagoras’ Theorem can actually be avoided even in Primary (Elementary) School problems.  So a 10 year old kid with a rudimentary knowledge of algebra could do this.  Can you spot a short cut?

Making Observations

     Stare at the diagram for a while.  What do you observe?

Solution

             area of  DDB_nC_n =  ½  of the area of  AB_nC_nD.

              area of  DDB_nC_n =  ½  of the area of  DB_nPQ.

        \  area of DB_nPQ  =  area of AB_nC_nD = n cm².

HeuristicsUsed

H04. Look for pattern(s)

H05. Work backwards

H09. Restate the problem in another way

H11. Solve part of the problem

Suitable Levels

* Lower Secondary Mathematics

* challenge for Primary school Olympiad

* other syllabuses that involve areas and a tiny bit of algebra

* anyone game itching for a challenge

[S2_20151122IXBQ] A Bonus for your Index Fun?

Question

Introduction

     This is a “bonus” question which most likely came from an Integrated Programme (IP) school.  It really tests the wits of students’ knowledge of indices and problem solving tactics.  I present two solutions without the use of logarithms.  The first solution uses reciprocal indices and the matching of the base ^a/_b.  In the second solution, we match up the indices to  xy.

Review of Important Laws of Indices

Solution 1

 

Solution 2

Remarks

     No logs from any forest were harmed in the process of making this blog post.  J

Heuristics Used

H04. Look for pattern(s)

H05. Work backwards

H09. Restate the problem in another way

H13* Use Equation / write a Mathematical Sentence

Suitable Levels

* challenge for Lower Secondary Mathematics (Secondary 2)

* GCE ‘O’ Level “Elementary” Mathematics (revision)

* other syllabuses that involve algebra and indices