[Pri5 20170426FEM] Baking Éclairs and Macaroons

Problem / Question

     This problem for primary 5 from one of my acquaintances on Facebook, considered to be of intermediate level difficulty (in Singapore).  But it looks rather challenging to draw all those bar diagrams, doesn’t it?

     Here is my quickie solution without explicit algebra and without bar diagrams.

Solution

     For convenience, we use 6 circle units for Eclairs and 6 square units for Macaroons.

Suppose there were half as many Eclairs and Macaroons, then there would be 15 more Eclairs.  So 3 square units add 15 can be changed to 3 circle units.

     Add 15 to the 17 and change 3 square units to 2 circle units.  We deduce that 5 circle units is the same as 85.  From here we can easily figure out the rest.

Ans: 102 éclairs.

Comment

     The problem can be solved by bar diagrams.  However, there are many ways to skin the cat.  For more good stuff, please join my Facebook group “Effective and Elegant Mathematics”.

Heuristics Used

H02. Use a diagram / model

H05. Work backwards

H06. Use before-after concept

H08. Make suppositions

H09. Restate the problem in another way

H11. Solve part of the problem

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_20151130IAXS] A Motif for the Absolutely Absolute

Problem

Introduction

     This question would pose a challenge for many students, although theoretically it is within reach of a good Additional Mathematics student (~ grade 10).  Graphs of both  sin x  and  cos x  are waves that oscillate up and down.  There are many pairs of vertical bars, indicating the absolute values or modulus, and these seem confusing.

Strategy

     Let us graph the functions  y = |cos x|   and  y = |sin x|.   Note that  ||sin x| – |cos x|| = ||cos x| – |sin x||.   The absolute difference of  |cos x|   and  |sin x|  is the difference between them ignoring the negative sign (if any) of the result.  And this is just the difference between the higher value and the lower value. 

Can you see any repeating patterns?  [H04]  Can you visualise the required area?  How many times is that of the basic pattern (known as “motif” in art)?  [H09, H10, H11]

Solution

Remarks

     Our total area is made up of four congruent pieces.  When  0 < x < ^p/₄,  cos x  is higher than  sin x.  That allows us to strip away all the absolute signs and do the calculation.

Heuristics Used

H04. Look for pattern(s)

H09. Restate the problem in another way

H10. Simplify the problem

H11. Solve part of the problem

Suitable Levels

* challenge for GCE ‘O’ Additional Mathematics  * IB Mathematics SL HL

* GCE ‘A’ Level H2 Mathematics  * IB Mathematics HL

* AP Calculus AB & BC

* University / College calculus

* other syllabuses that involve integration and area

* whoever is game for a challenge in integration

[Pri20151116DNGC] Quadratic Plays Second Fiddle in Product-Difference Riddle

Question

What two numbers give a product of  21.5  and a difference of  6.1?

Introduction

     This question reminded me of a question that I set myself when I was in Primary 4 (» grade 4).  I imagined a rectangle with breadth 4  and  length 2 units longer then the breadth (i.e. 6) giving a area (product) of  24.  Then I pretended that I did not know the breadth and let it be  x.  This led to a quadratic equation  x(x + 2) which I did not know how to solve (if I did not know the answer).  So I accidentally discovered quadratic equations when I was in Primary 4.  This led me to a quest to learn the method of factorisation (by “trial and error” or “guess and check”) and the quadratic formula.  I never liked trial and error.  So I continued in my quest to invent a method of factorisation that did not require “guess and check”.  I finally succeeded doing that in secondary 1 (» grade 7).  This turned out to be a Pyrrhic victory.  The method I invented was quite similar to the quadratic formula.

     There is a place for “guess and check” in mathematics.  I present a simple solution to the above problem using just that.

Solution

smaller #	larger #	product
2	8.1	16.2	û
3	9.1	27.3	û
2.5	8.6	21.5	ü

Solved! J

Heuristics Used

H02. Use a diagram / model    (table)

H05. Work backwards             (if the smaller number is this, what is the bigger number?)

H07. Use guess and check

H09. Restate the problem in another way      (area = product)

Suitable Levels

* Primary School Mathematics

* other syllabuses that involve decimal numbers

* anyone who loves to exercise their minds

[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

[H2_SAJC2006PromoQ1] Skipping Terms in an Arithmetic Progression

Question

Introduction

     This difficult-looking question has become pretty standard already.  There are some principles that the schools may or may not teach explicitly, but they expect students to know.  Let us review some of these principles.

Reminders

Refer also to this article.

Solution

Summary

     Remember that when you apply a formula (e.g. like the formula for the sum of an AP), you need to apply it with the appropriate numbers substituted.  Do not get stuck with the letters.  They are not meant to be taken literally, but change according to the situation.  For example, the “d” in the later part is  4  but it is different than the  d = 2  in the earlier part.

Heuristics Used

H04. Look for pattern(s)

H09. Restate the problem in another way

H10. Simplify the problem

H11. Solve part of the problem

H12* Think of a related problem

H13* Use Equation / write a Mathematical Sentence

Suitable Levels

* GCE ‘A’ Levels, H2 Mathematics 

* International Baccalaureate Mathematics 

* other syllabuses that involve arithmetic and geometric progressions

[S2 Expository] Negation and Swapping a difference

Introduction

     In this article, I introduce a neat and very useful trick for algebra and arithmetic.  Let us say you want to buy something that costs  $5,  but you only have  $2.  You need to borrow from your friend to help pay for it.  How much do you owe?  $3  Obviously.  Using negative numbers to denote debts, we can write out the story as  2 – 5 = -(5 – 2) = -3.  It is a ‘-’ because you know you are owing something.  To calculate the amount you owe, you just swap the  2  and the  5.

     In other words, you are instinctively using the following identity without realising it.                                             

This algebraic identity (something that is always true, not just sometimes true) actually works for all types numbers that the student will encounter, including positive and negative numbers.  It basically says that swapping a difference is the same as negating it. 

     How do we know it is always true?  This is easily shown:-

                      RHS = -(b – a) = -b + a = a + (-b) = a – b = LHS

The equality relationship is symmetrical.  Since  RHS = LHS,  we have  LHS = RHS.  Equations work both ways.

How to use it?

     (1) You can use it for calculations involving positive and negative numbers, as above.

     (2) Whenever you see difference e.g.  pet – cat,  you can swap the quantities, put brackets around it and attach a negative sign in front to get  -(cat – pet).  Remember that equations work both ways.  You can use it in the other direction.  Hence

     (3) Whenever you see a negated difference e.g.  -(train – van),  you can imagine that the negative sign causes the two quanties to swap.  After swapping, the ‘-’  is used up and you end up with  van – train.
     May the swapping power be with you!

Suitable Levels

* Lower Secondary Mathematics

* other syllabuses that involve algebra and negative numbers

[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

[AM_20150413RSD] Rationalising Denominators for #Surds

Question 

Introduction

     Expressions with surds in their denominators are cumbersome.  The good news is that we can make the denominators into rational numbers, which are nicer.  Rational numbers those that can be expressed as a ratio of integers i.e. they are (proper or improper) fractions or can be converted to fractions.  Whole numbers are also part of rational numbers because you can always put them upon a denominator of  1;  e.g. 2 = ²/₁,  so  2  is a rational number.

     The standard trick for simplifying expressions with surds in their denominators is to rationalise the denominator by mutiplying the numerator and the denominator with its conjugate surd.  For example, the conjugate surd of   Ö5 + Ö2   is   Ö5 – Ö2.   Just change the  +  to  –  or the  –  to  +.  Let us see how the magic works.

Solution

Remarks

     Note that in the first step, I pulled out 2 as the common factor of the denominator, so that I get a simpler surd to work with.  Always try to work with simpler expressions.  This not only shortens your working, it reduces your chances of making a careless mistake.
     In mathematics, “rationalising” does not mean you give some reason or excuse for something that you know you have done wrong.  It means “make it into a rational number”.  Why does rationalising the denominator work?  This is because on the bottom (denominator) we have a difference-of-squares expression of the form
                                           (a + b)(a – b)   which is equal to   a² – b².
Since squaring “gets rid” of square roots,  a²  and  b²  will give you rational numbers (whole numbers or fractions), you will end up with a nice number downstairs (on the denominator).  Pupils should make sure they have this technique in their repertoire of skills.

Suitable Levels
* GCE ‘O’ Level Additional Mathematics
* revision for GCE ‘A’ Level H2 Mathematics
* revision for IB Mathematics HL / SL
* other syllabuses that involve surds
* precocious kids who always want to learn more