## Wednesday, April 26, 2017

### [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”.

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

## Sunday, January 1, 2017

### [Enrich20170101SQT] Calculating Square Roots by Hand

Introduction
Happy New Year to our readers!  I wish this year will be a fruitful one for everybody.
Today, I will illustrate how to calculate square roots by hand, using  54 756  as an example.  It is similar to long division, but has some modifications.

Solution

Starting from the right, pair up the digits.

2×2 = 4  is the nearest perfect square to  5.  Subtract and bring down the next two digits, giving  147.

Double the digit  2  to get  4.  Think:  ? × 4?  gives  147  or nearest possible value.  We have 3×43 = 129.

Subtracting and bringing down the next two digits gives  1856.  Replicate the digit  4  on the left and double the digit  3,  giving  46.

Now think:  ? × 46?  gives  1856  or nearest possible value.  It turns out that  4 × 464 gives exactly  1856.  We are done!  The square root of  54 756  is  234.

How does it work?

This relies on the algebraic identity  (10a + b)² = 100a² + 20ab + b², the right-hand expression is equal to   100a² + (20a + b)b.  For example, at stage 4, we have  a = 23,  b = 4  and  (20a + b) = 464.
Did you learn something today?

## Sunday, November 6, 2016

### [AM_20161105ITFF] False Friends in Integration (Calculus)

Question

Introduction
False friends are words in two languages that look/sound alike, but differ significantly in meaning.  Do you know that there are also false friends in mathematics?  Can you distinguish and explain the difference between the two integrals?

Solution

The integrand on the left has the variable  x  as the base and the constant  e  as the index.  So we integrate it using the Power Law.
By contrast, for the integrand on the right, the base  e  is a constant whereas the index is the variable  x.  Integrating  e  to the power of  x  is the eeeeeeeeeeeeeeeasiest.  You just get back the same thing, plus the arbitrary constant of course.

Remark
Many students make the mistake of trying to apply the Power Law for the exponential.  As a learner of mathematics, one needs to cultivate the habit of being observant and paying attention to detail.  This is part of developing one’s identity and character which is important in life.

Suitable Levels
GCE ‘O’ Level Additional Mathematics
GCE ‘A’ Levels (revision)
* revision for IB Mathematics HL & SL (revision)
* Advanced Placement (AP) Calculus AB & BC
* University / College Calculus
* other syllabuses that involve integral calculus

* whoever is interested

## Thursday, November 3, 2016

### [Enrich20161103NRP] The Napkin Ring Problem

Problem

Two rings are made by drilling a cylindrical hole through a small sphere and a hole through the large sphere, such that the resulting rings have the same height (2h).
Which ring has the larger volume of remaining material?

Solution
The answer is: both rings have the same volume.  How can we know?
There is a way to show this using integration.  But calculus is not necessary.

Let  r  be the radius of any chosen sphere and let  a  be the radius of the cylindrical hole.  By Pythagoras’ Theorem,  h² = r² – a².  Consider a cross-section of the ring sliced a distance  x  from the centre of the sphere, perpendicular to the axis of the cylindrical hole.  The outer radius of this cross section is the square root of  r² – x².  Hence the area of the material in the cross-section is
p [(r² – x²) – a²]  =  p (r² – a² – x²)  =  p (h² – x²)
Note that  r  does not appear in the formula.  That means the cross-section does not depend on  rA bigger (or smaller) sphere would have the same cross-sectional area for each distance  x  away from the centre.  By Cavalieri'sPrinciple, the other sphere will have the same volume!
By the way what is this volume?  It is the same as that of a sphere without hole  i.e.  where  a = 0  and  r = h.  This works out to be  4/3 p h³,  where  h  is half the height of the ring.

H02. Use a diagram / model
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 (Number patterns, with algebra)
* revision for IB Mathematics HL & SL
* Advanced Placement (AP) Calculus AB & BC
* University / College Calculus
* other syllabuses that involve volumes and Pythagoras’ Theorem
* any learner who is interested

## Saturday, October 22, 2016

### [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.

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

## Wednesday, June 8, 2016

### The passing of a Great Giant - Jerome Bruner

Jerome Bruner, a great psychologist, has died recently.  We thank him for his theories that guided the development of Singapore Mathematics education.

One of his greatest theories that is useful for children's learning is the Concrete-Pictorial-Abstract (Enactive-Iconic-Symbolic) approach.

It is impossible to use only words to explain simple mathematical concepts to a young child, since those concepts cannot be further explained using words.  You will run out of words to explain!  What definitely does not work is to start from use just words and they don't "get it", then scold them for being stupid.

They have to learn buy touching and playing with things, then from looking at pictures and then using abstract words or reasoning.

## Saturday, February 27, 2016

### [S1_20160227FZCK] Factorisation by Chunking

Problem / Question

Solution

Commentary
Here I illustrate the usefulness of chunking to factorise (AmE: factor) an algebraic expression.  Observe that  3a – 2b  is a repeated part of the expression.  I call it a “chunk”.  To make it clear, I rewrite  (3a – 2b)²  as   (3a – 2b)(3a – 2b)  so that you can see it as two copies of the same chunk.  I highlight in yellow one copy of  (3a – 2b)  from each of
(3a – 2b) (3a – 2b)   and  -3(3a – 2b).  The remaining stuff are highlighted in blue and green.  Take out the yellow chunk as common factor by writing it out on the left in the third line, shown in yellow.  You can pull out the common factor by writing it out to the right if you want, but here I chose to put it on the left.  The result would be equivalent anyway.  Once you have written out the common factor,  you write out the other stuff (shown highlighted in blue and green) into another other bracket.
Once you understand how it works, you can actually do the second line mentally and write down the answer straightaway.  Chunking is a very useful technique in mathematics.  Here are some more examples of the technique of chunking: (1), (2), (3).

H04. Look for pattern(s)
H10. Simplify the problem
H11. Solve part of the problem

Suitable Levels
Lower Secondary Mathematics (Sec 1 ~ grade 7)
GCE ‘O’ Level “Elementary” Mathematics
* other syllabuses that involve algebra and factorisation (factoring)
* any learner who is interested