[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