[S2_20150523XFDS] Numbers that can be Difference of Squares

Question

Introduction

     This is likely an primary mathematics olympiad-type of question, but lower secondary pupils can also try this.  It involves deeper thinking.  But where do we begin?  Sometimes it is good to begin from the beginning, and then follow your nose. 

Reminders

Solution

     Suppose  N  is a whole number such that  1 < N < 1000  and  N  can be expressed as

                                        N = a² – b²  = (a – b)(a + b)

a difference of squares.  So  N  can be split as a product of two factors  (a + b)  and  (a – b).  Observe that     (a + b) – (a – b) = 2b,       which is an even number.

     The difference between the two factors is an even number.  This can only mean that the two factors are  both odd  or  both even.  You cannot have one of them odd and the other even, because when you subtract them, you would get an odd number.  We now have three cases:-

     Case 1a:  N  is even but not divisible by 4.

     Case 1b:  N  is divisible by 4 (and, of course, is even)

     Case 2:    N  is odd  i.e. both  (a + b)  and  (a – b)  are odd

Ans:  750

Remarks

     In the foregoing, it is possible for  b  to be zero.  0 happens to be a perfect square, because  0² = 0.  However, we need not worry about this, because the above algebra is general enough to cover the case where  b  is  0.
   We have solved the problem using logic, even-vs-odd analysis and the three important algebraic identities under reminders (highlighted in orange).  We also used the special cases (highlighted in light blue) and made observations based on them.

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
* Primary School Mathematics Olympiad
* Secondary 2 Mathematics » grade 8 (expansion and factorisation)
* anyone who is game for a challenge in algebra and number theory

[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 Expository] Square-of-Difference Identity for Algebra

     An algebraic identity is an equation that is true for all values of the variables involved.  If we substituted any set of values to the Left Hand Side (LHS) and the same values to the Right Hand Side (RHS), the equation will be true i.e. the LHS will always be equal to the RHS.  The square-of-difference identity

                                       

is one of the three identities that students have to learn in secondary two.  Many students have difficulty remembering this, and they mix this up with the other identity, which involves  a² – b².  However  (a – b)²  is not the same as  a² – b².  They do not understand why the above formula is true, because almost nobody explains it.  Perhaps a few teachers explain the identity for  (a + b)².  But if the  ‘+’  is changed to a  ‘–’  this is a little trickier.  Let me try to explain the formula visually, and with colours to boot, for perhaps the first time in history.

     We start (on the left) with a square of side  a,  whose area is  a².  This is shown in green in the diagram.  We partition each side of the square into  a – b  and  b.  Our goal is to get an area of   (a – b)².  Let us flip the strip of width  b  on the right of the square.  This strip has area  ab  and is shown in pink in the middle square.  This is the same as saying we are subtracting one copy of  ab.  Note on the bottom of the square, there is another strip of area  ab  (shown outlined in orange).  If we subtracted that, we would have subtracted  2ab  (see the square on the right), and we would seem to get  (a – b)².   But then the little square of area  b²  (indicated by a darker green) would have been subtracted twice.  So we need to add  b²  back, so as to restore balance in the universe. 

     You can imagine doing this with a square of area  a²  made of layer of sand.  We remove strips of area  ab  two times – from the right and from the bottom.  Then we patch up the  b²  hole by adding back a layer of sand.  We finally end up with a layer of sand of area  (a – b)².  This illustrates why  a² – 2ab + b² = (a – b)².

     Isn’t this kewl?

Suitable Levels

* Lower Secondary Mathematics

* other syllabuses that involve algebra, expansion and factorisation

[Pri20150429PPP] Pythagoras for Primary Pupils?

Question

Introduction

     This primary (elementary) school mathematics examination question created quite a stir among some parent support groups on Facebook.  The issue is that the height of the triangle seems to have been omitted.

Pythagoras’ Theorem

     Some participants who know secondary school mathematics were quick to suggest the use of Pythagoras’ Theorem to find the height of the triangle, which works out to be  12 cm.  This leads to the answer (2) 114 cm²,  which is correct.  The problem is that pupils are not taught Pythagoras’ Theorem until secondary school, and so it would seem an unfair test for the pupils.  So the discussion turned to thinking of various methods by which a primary school pupil may uncover the answer without resorting to advanced knowledge.

Elimination and Educated Guessing

     Mr Teo Kai Meng, a tutor who regularly participates in the support groups, offered some insightful observations.  Assuming that the height measurement is a whole number of centimetres, only options (2) and (4) need to be considered as they were divisible by  19, which the area had to be under the said assumption.  [ Another tutor, Melissa Song had a similar idea by observing that since the triangles DKLN and DKMN  have the same height, the ratio of their areas is the ratio of their bases LN : MN = 19 : 16. ]  We can ignore choices (1) and (3).  As we know,  area = ½ ´ base ´ height .  Since teachers like to catch students for being careless in forgetting to multiply by ½  (or dividing by 2), it is quite likely that option (4) was set up as a booby trap.  Thus one may intelligently surmise that option (2) should be the answer.

     Mr K L Chua, a tutor who calls himself “Mathematics Specialist”, used a similar reasoning.  He worked backwards from each of the four choices to get the heights and chose the most plausible answer.  Of the two whole-number answers, (4) was eliminated and (2) was chosen since from the diagram the height should roughly be near to  16 cm  even though the diagram was not drawn to scale.

Scale Drawing

     Another participant suggested doing a scale drawing to estimate the height.  Indeed this can be done, and is a good tactic too, since this could be done quickly with a ruler and pencil.

A Visual Solution

     Assuming no knowledge of Pythagoras’ Theorem, it is possible to construct a visual solution.  First we take four copies (indicated in orange/light-orange) of the right-angled triangle  DLNK  and arrange them to surround a square of side  37  cm (indicated in green),  which is the same as the longest side (hypotenuse) of the said triangle.     This green area is  1 369 cm².  Now we rearrange the triangular pieces as indicated by the red arrows.  The areas of the orangey triangles do not change when you shift them.  Neither will the green area change, since the total area everything in the containing square (orangey plus green areas) remains the same.

With pairs of right-angled triangles joined together along their longest sides, we now obtain two green squares, the larger of which has side  35 cm.  This gives an area of  1 225 cm².  The total area of the two green squares is the same as the area of the large green sqaure before the shifting, namely  1 369 cm².  Hence the area of the small square is  (1 369 – 1 225) cm² = 144 cm².

From here we quickly deduce that the unknown height is  12 cm,  12 being the square root of  144.  Hence we conclude that the area of  DKLM  is  114 cm².

Remarks

     I have shown that it is theoretically possible for primary school pupils without knowledge of Pythagoras’ Theorem to derive the answer in an exact manner.  By the way, the method of shifting triangular pieces as indicated above can be generalised to give a proof of Pythagoras’ Theorem.

     Some parents expressed fear that this is another one of those Cheryl-like or olympiad type of problems.  Is there a conspiracy by the school teachers to purposely set difficult questions and make life difficult for pupils, disadvantaging those who cannot afford private tutors?  In this case, could this just have been an oversight on the part of the teacher who set the question?

     Entrepreneur John Low Jiayong and tutor John Lim sourced for and managed to obtain faithful copies of the original question.  It turned out that some school-paper vendors had inadvertently erased the 12 cm measurement.  Thus in the original question, the height  KN  was given as 12 cm, and the measurement of  37 cm for the hypothenuse  LK  was purposely given as extraneous information to distract students.

     Some parents observed that this is after all just a multiple-choice question that carries a credit of only one mark.  If this were an exam situation with the  12 cm omitted, it would be best to either sacrifice the  1  mark and move on, or to use tactics like elimination, educated guessing or estimation with scale drawing.
     Anyway, it had been quite a fruitful discussion, as adults (parents and tutors) attacked this problem from many angles in a purposeful way.  Many parties put in concerted effort and contributed in an engaged matter.  It would be good if this type of rich discussion were enacted in our classrooms everyday among pupils and teachers, perhaps enabled by social technology.  For then, pupils would deeply learn and perhaps would not be so stressed out, nor be needing so much extra external help.  Private tutors could then move on to focus on value-added  mentorship  for 21^st Century Learning, instead picking up the tab where school teachers have left off.

Heuristics Used / that can be Used

H01. Act it out (e.g. scale drawing)

         (as a class learning activity, pupils can use scissors to cut out paper triangles

          and physically move them around)
H02. Use a diagram / model

H04. Look for pattern(s)

H05. Work backwards

H07. Use guess and check

H08. Make suppositions

H09. Restate the problem in another way

Suitable Levels
* Primary School Mathematics
* Lower Secondary Mathematics
* any precocious / gifted pupil who wants to learn
* any person of any age interested in creative problem solving