[Pri20150402PTSMAP] A Staircase with Higher Steps

Question

Introduction

     This pertains to the sum of consecutive numbers with constant skips.  I set this question

to illustrate the heuristic of looking for patterns [H04].  It is similar to this question, except that now the numbers jump or skip by  2  instead of just   1.  The more knowledgeable reader will doubtless recognise this to be an arithmetic progression.  The challenge now is how can a primary school pupil do it without having learnt about any more advanced mathematics or algebra, relying purely on pattern recognition.

Solution

     As in the previous solution, imagine the sum as a series of vertical bars.  The numbers all jump by  2  this time.  Because the jump amount  2  is constant, you see a nice staircase pattern (shown in violet).  Each step of the staircase is of height  2  units.  If we make a copy of it and turn it upside-down (shown in green), the two staircases join together nicely to form a rectangle.  Notice that  101+3 = 99+5 = 97+7 = ... etc and they are all equal to  104.  If we know the number of columns, we can work out our desired sum.  How many columns are there?

     The number of columns is the same as the number of terms in  our sum.  OK, but then how many terms are there?  How to calculate this?  Let us look at a few simple cases first [H10. Simplify the problem].

Let us try to observe the pattern.  Note that the size of each skip is always  2.  If there are  2  terms, it is just  3  and  5,  there is one skip of  2.  From  3  to  7,  there are  3  terms, there are two skips of  2  each.  From  3  to  9,  there are  4  terms,  the difference is  6  and there are  3  skips.  From  3  to  11,  there are  5  terms,  the difference is  8  and there are  4  skips.  If you go from  3  to  13,  the net jump is  10  and there are  5  skips  and  6  terms.  We can tabulate the data into a table [H02] below:-

        skip size = 2

Start	End	Total Skip	# skips	# terms
3	5	5 – 3 = 2	2 ¸ 2 = 1	2
3	7	7 – 3 = 4	4 ¸ 2 = 2	3
3	9	9 – 3 = 6	6 ¸ 2 = 3	4
3	11	11 – 3 = 8	8 ¸ 2 = 4	5
3	13	13 – 3 = 10	10 ¸ 2 = 5	6

Do you notice some things?  [H04]

The total skip is the difference between the starting and ending numbers.

The number of skips is the difference divided by the skip size.

The number of terms is always one more than the number of skips.

Since our last term is  103,  the total skip is  101 – 3 = 98.  The number of skips is  98 ¸ 2 = 49.   So there are  50 terms  i.e.  50  columns.

Hence the size of our rectangle is  50 × 104.  But we only want half of this rectangle (shown in violet).   Hence the sum is  ½ × 50 × 104 = 2 600.

Ans:   3 + 5 + 7 + ... + 99 + 101 = 2 600

Summary

     This article illustrates the heuristic [H04 Look for pattern(s)].  Our first pattern we notice is the staircase pattern.  After making a copy and turning that around, we notice that it forms a rectangle, with columns of size  104  each.  Now we look for a pattern that enables us to find the number of columns, which is the number of terms in our sum.  We note that the number of terms is always the same as the number of skips, which is the same as the difference between the start and the end all divided by the skip size.  This enables us to solve the challenge in a way similar to my previous example.

Reflections
     Do you think this method will work for different starting numbers and different ending numbers?  For different skip sizes?  Why not set up your own similar question and try it yourself and see whether it works?

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

* GCE ‘O’ Level “Elementary” Mathematics (Number patterns, with algebra)

* GCE ‘A’ Levels H2 Mathematics (sequences and series, with algebra)

* IB Mathematics (sequences and series, with algebra)

* anyone who loves patterns and relishes a challenge

[PriCNPTGN_20150508] Sum of the First Few Natural Numbers

Question

 Introduction

     Numerical patterns are a challenge to learn and to teach.  Those of us who are teachers, usually the better students of our time, tend to think that the pattern is “obvious”, and hence do not bother to explain and/or facilitate classroom discussions regarding patterns.

     This article is about a pattern that involves the sum of the first few natural numbers (or wholenumbers).  This pattern, common in our primary school mathematics patterns, is a special case of the sum of an arithmetic progression, but its formula is not usually taught explicitly.  The story is told of the great mathematician Karl Friedrich Gauss, who figuredout a short-cut for adding up  1 + 2 +... + 100.  Lesser mortals in primary schools are left to struggle with frustration, or to copy “model answers” from their tutors or teachers without understanding how the solutions were obtained.
     As in a previous article, I attempt to illustrate the pattern visually.  I shall highlight the linkages to facilitate discovery of the general method, as well as show how the last part (part(c)) can be solved.

Solution

     Many pupils are able to deduce the answer to part (a) by analysing the differences between successive answers, which is equivalent to asking what must you add to get the next number.  For example, from the 1st number  1,  you add  2  to get the 2nd number  3,  you add  3  to get the third number  6.  To get the 4th number, you add  4  to get  10.  This approach works, but will not help you much for parts (b) and (c) of the question.  The better way is to look for a method that does not require you to keep on adding numbers.  That would allow one to kill all the birds with one stone.  How to do this?

     Imagine the given dots being doubled, rotated and then put together into parellelogram-like matrices.  If you want, you can imagine them as rectangular arrays.  I use the colour orange for the originals, and blue for the copies.

     As you can see, for figure 1, the total number of blue and orange dots is  1 ´ 2.  For figure 2, the total is  2 ´ 3.  For figure 3, the total is  3 ´ 4.  For figure 4, the total is  4 ´ 5.  The pattern is: for whatever number representing the position of the figure, the total is this ordinal number multiplied by another factor that is one more than this number.  Hence the number of original dots (shown in orange) is this product divided by  2.  With this insight, we can fill up the table to answer part (a).  The answer for figure  5  is  30.

We can also figure out that the answer to part (b) is  55,  with the above-mentioned pattern.

For part (c), we can use trial and error or “guess and check”.  Trying  21  gives  21´22 = 462  which does not work.  We try 22:  22´23 = 506.   Yes!  Bingo!

You may also use a calculator to help you.  Since the two unknown factors are close together (they differ only by one), it is almost like multiplying a number with itself, or squaring.  So to guess our number, we may use the square root (the opposite of squaring) to estimate it.  The square root of  506  is about  22.49.  We guess  22  and verify that 22´23 = 506.

Summary

     The pattern involving the sum of the first few whole numbers may be deduced by making a copy of the original figure, rotating it and joining it to form an array.  Just multiply accordingly and then divide by  2  to get the sum for the original figure.  To solve for a figure’s ordinal number (which figure has a certain given number of dots), one may use guess and check, or use square roots.  A primary school pupil should be able to all these without the knowledge of advanced techniques like the sum of arithmetic progressions or quadratic equations.

Heuristics Used

H02. Use a diagram / model

H03. Make a systematic list

H04. Look for pattern(s)

H05. Work backwards

H07. Use guess and check

H09. Restate the problem in another way

Suitable Levels

* Primary School Mathematics (algebra unnecessary)

* GCE ‘O’ Level “Elementary” Mathematics (Number patterns, with algebra)

* GCE ‘A’ Levels H2 Mathematics (Number patterns, with algebra)
* anyone who loves a challenge to unravel a pattern

[S2_ABNPQX_20150502] Jack-in-the-Box of Fractions

Question

Introduction

     This number pattern question requires a lot of observation and algebra skills.  One also needs to understand how subscript notation like  F_n  works.

Solution

[H2PCRP20150406] Colouring The Pentagon (Combinatorics)

Introduction

     This is a question suitable for the mainstream Junior College students taking H2 mathematics, but is some primary school olympiad question from somewhere.  Whatever!  Mathematics is for everybody, young and old.  Anybody can solve this problem if  s/he makes observations and uses the right approach and thinking skills.

An Incisive Insight

     Although this pentagon is not a regular pentagon, the colouring scheme depends just on the order of colours on the edges.  We can start from one edge and see what colours are possible.  And then we can rotate the colouring scheme around.  [ We are breaking down and simplifying the problem. ] 

     Fiddling around with various possibilities, you might realise that:-

·  you cannot have three of the same colour going round the pentagon

·  you cannot have three sets of pairs of edges with the same colour.

·  you cannot two single colours and one double colour

There must be one single colour and two pairs of doubled colours.  All colouring schemes will have a  “12123”  colouring pattern going around in a loop.  The diagram below shows an example where colour 1 = yellow (Y), colour 2 = red (R)  and colour 3 = blue (B).

Do we need to consider a “21213”  pattern?  If  “12123”= “YRYRB”, we can later reassign colours, swapping R and Y to give 1=R and 2=Y and then “12123”=“RYRYB”.  So we have got that covered.  Let us worry about the reassignment later. 

Solution

     Observe that the position of the “3” (the single colour) can be rotated round the edges of the pentagon in  5  ways.

     Observe also that there are  3! = 3 ´ 2 ´ 1 = 6  ways to shuffle the colours i.e. assign colours 1, 2 and 3  to  Y, R and B.

The above two processes (rotation and shuffling) are independent of each other.  Rotation of the single colour can be done with or without the shuffling of colours.  Hence we can use the Multiplication Principle and calculate

          the total number of ways = 5 ´ 6 = 30.

Tada!

Heuristics Used

H02. Use a diagram / model

H04. Look for pattern(s)

H09. Restate the problem in another way

H10. Simplify the problem
H11. Solve part of the problem

Suitable Levels

* GCE ‘A’ Level H2 Mathematics

* IB Mathematics HL / SL

* Primary School Maths Olympiad

* other syllabuses that include combinatorics

[JCH2CNAGDH_20150402] Vector Rotation via Complex Numbers

Question

The points  D  and  G  in an Argand diagram represent the complex numbers  3i  and  6 + 7i  respectively.  DG  is in fact a diagonal of the square  DFGH.  Find the complex numbers represented by  F  and  H.

Introduction

     This is a question that tests students’ concept of the geometry complex numbers.  The key to solving this question is to understand that multiplying a complex number  re^iq  gives a stretching effect, as well as a rotating effect.  The stretch is by  r  times  and the rotation is by  q  radians anti-clockwise.  If  r = 1,  then there is just rotation and effectively no stretching.

Important Principles

Solution

[Pri20150330FBT] Open-ended Question on #Fractions

Question

Give a fraction with its value in between 2/5 and 1/2.  (Note: the denominator cannot be greater than 20)

Discussion

This is an open-ended question for primary (elementary) school, which ought to be easy because there are many possible answers and you just need to supply one that meets the requirements.  However, most people are used to closed-ended questions, which have only one correct answer.  This is what prompted a parent to pose this question on a parent-support forum on Facebook. 

Many people, including parents and even some private tutors, began to supply their answers to this question.  Much of the discussion clustered around the idea of the mid-point or the average of the two given numbers, namely ¹/₂ (²/₅ + ¹/₂) or  ⁹/₂₀, and why that gives an answer.

Then Dr Kho Tek Hong (retired curriculum specialist, the “father” of Singapore mathematics) chipped in.  He said that there are many possible answers e.g. ³/₇, ⁴/₉, ⁵/₁₁, ⁵/₁₂, ⁶/₁₃, etc.  He suggests that students be asked to justify their answers.  [ By the way, here is an interesting method to compare fractions. ]

And I thought that was a stroke of educational brilliance from The Guru!  It shows the spirit of the Singapore mathematics curriculum – to get students to think, to be open-minded, to use logic and to be able to communicate mathematically.  The Singapore curriculum is not meant to torture students, nor to get them to toil around a tortuous path around a high mountain seeking the elusive holy grail – although it often seems that way.  Teachers, tutors and parents would do well to help the learners it in the right spirit.  Mathematics is not always about calculations (although you need to do some calculation).  We also need to open our minds to multiple answers as well as to embrace various methods and concepts.

[Pri20150330CPF] How to #Compare #Fractions

Comparing Positive Fractions

     How do you compare positive fractions, which are taught in primary (elementary) school?  For example, which is bigger:  ⁵/₆  or  ³/₄ ?

The “orthodox” method is to put them both to a common denominator.  The Lowest Common Multiple (LCM) of  6  and  4  is  12.  Multiplying the left fraction by  ²/₂  and the right fraction with  ³/₃  gives, respectively,  ¹⁰/₁₂  and  ⁹/₁₂.

Since  ¹⁰/₁₂  >  ⁹/₁₂,  we conclude that  ⁵/₆  >  ³/₄.

Another Method

     Here is a “short-cut” that I learned from a schoolmate in primary school.  Basically you “cross-multiply”: multiply the left numerator with the right denominator, and multiply the right numerator with the left denominator, and then compare the products so formed.  That will give you the correct inequality or equality sign (viz. ‘<’, ‘=’ or ‘>’).

As we can see, since  20 > 18,  we conclude that  ⁵/₆  >  ³/₄. 

     Does this method work?  Yes, definitely.  You can try it out with a few pairs of fractions and you can see for yourself that it is so.  Is this method legit?  Why does it work?    I give a formal proof of the method below.

Further Discussion

     Use the above method with care.  Some school teachers may not accept the method not because it is not correct, but it sounds “dubious” to them because they have not heard of it or they are not able to prove it for themselves.  Pupils can use this short-cut to give them a quick look-ahead to certain questions, and as a back-up to check their answer after using the Lowest Common Denominator method.  In questions that ask pupils to arrange a few fractions in ascending and descending order, this “crossing method” may give some speed advantage if done carefully.

     In primary school, pupils focus on positive fractions.  In secondary school, negative numbers and fractions are introduced.  Does the above trick work for negative fractions?  Was my theorem and proof above carefully phrased enough to cover the negative fractions? 

     Note that this method works for comparing two fractions at a time only.  Sometimes this cross-multiplying gives rather big numbers.  In that case, it is better to multiply each fractions by the LCM of their denominators.  Essentially this is the same as the orthodox method, except that we do not write the denominators.  Can the above proof be extended to cover this new short-cut?  What do you think?

[Pri20150303VSA] A way to Visualise Arithmetic Progressions

Question 

What is the sum of 25 + 26 + 27 + ......... + 189 ?

Introduction

     This is a problem meant to stretch the minds of Singapore primary (elementary) school pupils.  Adults (e.g. parents, teachers and tutors) trying to help out usually recommend doing this using the “rainbow” method (where a bunch of arcs are drawn joining 25 and 189, 26 and 188, 27 and 187 ... and so on, forming something that looks like rainbow), or the sum of arithmetic progression formula

          ½ ´ number of terms ´ (first term + last term)

which is usually taught at the junior college (pre-university) level.  Although correct, do the learners understand the logic behind them?

A Visual Method

     In my visual representation below, the answer pops out almost immediately and the reasoning is made apparent to the student.

     Imagine a series of vertical bars representing 25, 26, 27 ... up to 189 joined together forming a staircase (shown in orange).  Make a copy of this (shown in blue) and flip it around and join the two shapes together to form a rectangle.  Note that the width of this rectangle represents the number of terms 189 – 25 + 1 = 165, while the uniform height is in fact first term + last term = 189 + 25 = 214.  Taking the “area” of the rectangle and dividing by two, we get  17 655, an answer that would agree with those found using the previously mentioned methods.  The diagram above is a kind of “proof without words”.

[AJC Promo 2012 Q5] Range of Composite Functions with Inverse

Diagram 1.  Whole Question

     This is a question taken from Anderson Junior College, one of Singapore’s above-average junior colleges (in terms of the calibre of student intake).  By now this college is reputed to set the most difficult examination questions in Singapore.  It seems that they are trying to give the top junior colleges a run for their money, so to speak.  The question is difficult because it really tests students’ understanding of the concepts.  If you do not understand what is happening, you would be totally lost – even your graphing calculator (GC), the student’s favorite psychological crutch, would not be of much help.

Diagram 2.  Part (i) of Question

     Part (i) of the question tests students’ understanding of 1-to-1 functions (a.k.a. one-one functions or injective functions).  A function is invertible if and only if the function is one-one.  [ For the current A level H2 syllabus, it is assumed that the codomain is always the same as the range, so there is no need to worry about survjectivity. ]  Using the GC to graph the function  f   and to obtain the local maximum point, one sees that the required domain is  -2 < x < 0.  

Ans:  k = -2.   [ I am using black for explanations and blue for written answers. ]

     The domain is highlighted in yellow in the diagram below.  If the yellow region were to extend to the left beyond this point, it would be possible for a horizontal line to cross two points in the yellow region.  That would make the function  f  not 1-to-1 and hence not invertible.

Diagram 3.  Graph of f

Diagram 4.  Part (ii) of Question

For part (ii), we recall that a composite function exists if and only if the range of the first function (read from right to left) is a subset of the domain of the second function.
We require

Range of g  Í  Domain of f-1

The domain of  f^-1  (the second function) is actually the range of  f.  From the above diagram, the relevant part of  f(x)  goes from -¥  to  f(-2) = -2 + ln 4 = 2 ln 2 – 2 » -0.614.  The range of  g is everything from -1 downwards (see diagram below: imagine taking every possible point of  R, the domain of g on the x-axis and shooting them over to the y-axis).  We write

            Range of g     = (-¥, -1]
            Domain of f^-1 = (-¥, 2 ln 2 – 2]

Since Range of g Í Domain of f^-1,  therefore  f^-1g  exists.

Diagram 5.  Range of g

     Finding the range of composite functions is something that many students have difficulty with.  There are two methods: the direct method and the two-step method.  The direct method is usually difficult or infeasible.  In this case, finding range from the graph of  y = f^-1g(x) is practically impossible, because there is no simple formula for  f^-1.   

The two-step method: Step 1.  Find the range of the first function. Step 2.  Transfer this range to the x-axis of the graph of the second              function and map every point therein over to the y-axis.

     Step 1 has been done already.  We have found that  Range of g = (-¥, -1] . 

Diagram 6.  The Two-step Method

     For step 2, although the formula for  f^-1  is impossible to find (it’s a pretty nasty question, isn’t it?), we know that this graph is a reflection of the graph of  y = f(x)  (shown in blue on the diagram on the right) in the line  y = x, and we can sketch this (shown in red on the diagram on the right).  Now transfer the range of g from the y-axis of your first diagram over to the x-axis of this diagram on the right (shown in green).  Now imagine taking every possible point of this set and mapping it over to the y-axis of the second graph.  The problem is: how to find  f^-1(-1)  when we don’t even know the formula for f^-1?  (really evil problem, isn’t it?).  One way to deal with this is to make an educated guess for  f(what?) = -1. 

     Notice that f(-1) = -1.  Therefore  f^-1(-1) = -1.

     What if your intuition really sucks and you cannot make a guess?  The GC can come to your rescue.  Set up the graph of  y = f(x)  with the restricted domain and then intersect that with the graph of  y = -1.  The intesection is at  x = -1,  which means f(-1) = -1,  or  f^-1(-1) = -1,  as above.

Diagram 7.  Using Intersection on the GC

     Going back to diagram 6: From the range of g on the x-axis of the graph on the right, the points will land on every point from 0 down to f^-1(-1) = -1, including -1 but excluding 0 (because x-axis is an asymptote for  y = f^-1(x)).  We answer thus:-

     Range of  f^-1g = [-1, 0)

Diagram 8.  Part (iii) of Question

Part (iii) tests students understanding of increasing and decreasing functions.

f  is an increasing function means      whenever   a > b,   f(a) > f(b)   f  is an decreasing function means      whenever   a > b,   f(a) < f(b)

These are in fact the definitions of increasing and decreasing functions.  One can recognise an increasing function from its graph by the up slope (positive gradients) as you move from left to right.  For a decreasing function, the slope will be down as you move from left to right (negative gradients).  [An interesting note:  if  f  is a decreasing function,  then  f^-1  will also be a decreasing function.]

   In our case,  the graph of  f  is down-sloping, so it is a decreasing function. 

          Since  f  is a decreasing function,  whenever   a > b,   f(a) < f(b).

Once again we seem to have the pernicious problem of not knowing formula for  f^-1.  How to solve the inequality then?  Well, we can apply  f  to both sides of the inequality and the inequality reverses (because f is a decreasing function).  Note that  f  and  f^-1  “cancel”  as functions  i.e.  ff^-1(w) = w  for whatever the  w  is as long as it is well-defined.  Hence we proceed as follows

                                          f^-1g(x) > -1

                                        ff^-1g(x) < f(-1)

                                              g(x) < -1

                                          -1 - x² < -1

                                               - x² < 0

This latter inequality is the last trick on the question-setter sleeve desgined to unsettle the student.  How do you solve this inequality?  Do you need to equate or intersect with anything?  Anyway, what is the meaning of solve?  

To solve an inequality means to find all the possible values of  x  such that when you substitute each value into the inequality, the inequality becomes a true statement.

Note here that if we substituted  x = 0,  we would get  0 < 0, which is not true.  However, if we substituted any other real number,  x²  would always be a positive number, the LHS would always be a negative number, which is less than zero.  Conclusion:  x  can be any real number except 0.

Ans:  x Î R \ {0}

Note that the written solution (the parts typed in blue) is actually very short, although the explanation is rather long, because a lot of deep thinking is involved.

Reflection

Let us think back on the lessons learnt while solving this particularly difficult problem.

*  the reason why this problem seems difficult is because it tests students’ understanding of
    concepts (which most are weak in).  From experience with many cohorts of students, the
    JC teachers know what concepts students are weak in and they like to set questions that
    exploit the chinks in students armours.

*  remember the horizontal line test and domain restriction to get a 1-to-1 function, so that the
    function is invertible.

*  remember the condition for the existence of composite functions

*  inverse functions swap the domain and range with the original functions

*  the two-step method is recommended for finding range of composite functions

*  increasing functions preserve inequalities, while decreasing functions reverse
    inequalities.  You can recognise a decreasing function from the downward

    slope of its graph as you go from left to right.

Although you do not have the formula for  f^-1,

*  the value of  f^-1(-1)  can be found by intersecting the graph of  y = f(x)  with the horizontal line
     y = -1.

*  f  and  f^-1  “cancel” each other.

Finally,

*  what is the meaning of “solve an inequality”?

*  How to solve inequalities like  -x² < 0?  What about  -x² > 0?   x² > 0?