[S1_20151226NPGT] Finding the General Term of a Sequence (1)

Problem

Introduction

     This is a typical Secondary 1 type of problem involving number patterns.  Students are usually able to see the link between successive terms, but the general formula seems to be a challenge for most.

Strategy

     In case this is not obvious, every time you go to the next diagram, you add four dots on the outside.  So you can fill in the table very easily.  For diagram 5, there would be  19  dots and the total number of dots up to diagram  5  would be  55.

     What is the number of dots for diagram  1 000  or any number  n  for that matter?  Now, some students may have a problem predicting beyond the first few numbers.  What we need is a expression or formula that predicts the number of dots given the diagram number  n.  You know that the sequence  3,  7,  11,  15,  ...  follow a pattern where you keep adding  4.  Have you encountered a sequence in which  4  is added each time?  Yes!  It is the 4 times table.  Suppose we have the 4 times table.  [H08]   Let us do a comparison between that and  D_n.

diagram #	1	2	3	4	5	...	n
4 times table	4	8	12	16	20	...	4n
Dn	3	7	11	15	19	...	?

The numbers in  D_n  are always one less than those in the  4  times table.  So  D_n = 4n – 1.

Solution

n	number of dots for the  nth  diagrams Dn	Sum of number of dots for the first  n  diagrams Sn
1	3	3
2	7	10
3	11	21
4	15	36
5	19	55

     D_n = 4n – 1

     S_n  = n(2n + 1)   ©

Commentary

     How can we get the formula for  S_n?  We can do so by trying to factorise the numbers [H09], and then look for pattern.  [H04, H05]

           3 = 1×3   = 1×(2×1+1)

         10 = 2×5   = 2×(2×2+1)

         21 = 3×7   = 3×(2×3+1)

         36 = 4×9   = 4×(2×4+1)

         55 = 5×11 = 5×(2×5+1)

         ...

                      S_n = n(2n + 1)   ©  bingo!

But what if you have poor observational powers and if you are desperate?  There is a secret weapon to handle this!  Please refer to this article.

Heuristics Used

H04. Look for pattern(s)

H05. Work backwards

H08. Make suppositions

H09. Restate the problem in another way

Suitable Levels

* Primary School Mathematics (challenge)

* Lower Secondary Mathematics (Sec 1 ~ grade 7)

* GCE ‘O’ Level “Elementary” Mathematics (revision)

* other syllabuses that involve number patterns and algebra

* any precocious or independent learner who loves number patterns

[Pri_20151224RAMN] A Coin Problem with Constant Difference

Problem

Danny saved some  50-cent coins and  $1-coins in his coin box.  The total value of the  50-cent coins to the total value of the  $1-coins he had was in the ratio  2 : 5.  After  $14  worth of  50-cent coins and an equal value of  $1-coins were added to the coin box, the ratio of the total value of  50-cent coins to the total value of  $1-coins became  5 : 9.  How many coins of each type did Danny have in the end?

Introduction

     Here is a “Singapore math” coin problem that can be befuddling for kids and even for adults.  To rub salt to the wound (or pour oil to the fire?), the value of a collection of coins is different than its number.  Whilst a $1-coin obviously has a value of one dollar, you would need two 50-cent coins to make up a dollar.

Strategy

     Notice that after adding  $14  worth of coins to both types of coins,  the difference in the total value of the two types of coins remains the same.  Some people call this a “constant difference” problem.  But how do we exploit this constant difference, when the type of ratio units used in  2 : 5  are most likely not the same as those used in  5 : 9?  Well, we need to bring them to a common unit! [H09]   How?  Read on!

Solution   [H02, H06]

Ans: Danny had  54  $1-coins and  60  50¢-coins in the end.

Commentary

     I am using Distinguished Ratio Units in my presentation.  This makes it clear that the ratio units are of different types.  In the the “before” stage [H06], the difference in the value of the two sets of coins is  3  circle units.  In the the “after” stage, the difference in the value of the two sets of coins is  4  square units.  But we know these two differences refer to the same numerical number.  The Lowest Common Multiple of  3  and  4  is  12.  So both of them must me equal to  12  common units (which I envelop with triangles).  We multiply the numbers inside the circle units by  4  and we multiply the numbers in square units by  3.  I put these multiplications in quotation marks because we are not really changing the numbers of coins.  We are merely changing the type of units used.  I am saying that each square unit is the same as  3  triangle units and each circle unit is the same as  4  triangle units.

     Once we bring everything to common units (triangle units), we can see the  $14  added corresponds to  7  triangle units.  Henceforth the whole problem unravels easily.  [H11, H05]

Heuristics Used

H02. Use a diagram / model

H04. Look for pattern(s)

H05. Work backwards

H06. Use before-after concept

H09. Restate the problem in another way

H11. Solve part of the problem

Suitable Levels

* Primary School Mathematics

* other syllabuses that involve whole numbers and ratios

* any independent learner who is interested

[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

[S1_20150510AESFRN] Figuring out Different Significant Figures

Question

Vishal estimated a number as 2000 when he rounded the number to 1, 2 or 3 significant figures. Ethan was shocked by Vishal's answer and he asked Vishal what is the smallest and largest possible number such that the estimated number is 2000.  Can you help Vishal answer Ethan?

Introduction

     This is taken from ShingLee’s Secondary 1 Mathematics (7th edition) textbook.  It is a good question to test if students really understand the meaning of significant figures!  If you have not done so, please read my article Significant Figures Made Easy.

     To round certain number to  n  significant figures:-

Step 1:  Identify the first non-zero digit from the left. Step 2:  From this digit, start highlighting  n  significant figures. Step 3:  Look at the digit just after the last highlighted digit.  Round up if              this is 5 and above, otherwise do nothing. Always remember to include the place-holder ‘0’s.

Solution

     This question requires one to think backwards.  You are given the number  2000.  If it is correct to  1  significant figures, then what are the possible numbers that round to it?  If it is correct to  2  significant figures, then what possible numbers round to it?  If it is correct to  3  significant figures, what are the possible numbers round to that?  The question is not specific about what sort of “number” they meant, so I shall assume that they refer to real numbers.

Technical Remark

     In case you are wondering, for example,  why in the answer for  2  significant figures i.e. the interval
1 950 < x < 2 050, the  x cannot be made equal 2 050, this is because it would be rounded to  2 100.  A number like  2 049.9999999  (seven ‘9’s after the decimal point)  will be rounded down to  2 000.  Even if there were one trillion ‘9’s after the decimal point, as long as the number of decimals terminates (i.e. is finite), it will still be rounded down to  2 000.

     Once a number touches  2 050  exactly, it will be rounded up to  2 100  for  2  significant figures.  So technically, there is no largest number, because you can keep on adding more and more  9’s  behind as long as it is a finite number.  But you cannot add an infinite number of  9’s  because that would touch  2 050. 
2 050  is not the largest number, as it is not among the possible numbers, but  1 950  is the lowest number, because it is among all the possible numbers.   Because there is actually no largest number, I used the word “largest” in quotation marks.  [ Optional: The interested reader may want to check up infimum  and supremum ]  These are some very fine details that the textbook authors and editors forgot.  Even if the editor has a PhD, the PhD could be in “mathematics education” but not “mathematics”, and may have forgotten the rigours of the real numbers system.  The answers given at the back of the book are not only wrong, one pair of answers was missing.  This caused some confusion among parents and students.  I suspect they employed a non-professional part-timer to do up the answers in order to save costs.

Suitable Levels

* Lower Secondary Mathematics (usually secondary 1 ~ equivalent to about grade 7)

* other syllabuses that involve estimation and approximation

[S1_Expository] Significant Figures Made Easy

Introduction

     When you read a textbook on the topic of significant figures, you often get a bunch of complicated rules which seem pretty arbitrary and do not make sense.  Frankly, I too found these rules confusing.  They do not explain why those rules exist.  But there is sure-fire short-cut way to round a number to a given number of significant figures, based on understanding two key concepts: (1) the meaning of “significance” and (2) the role of ‘0’.  I present these below.

What does “significant” figure mean?

     The word “significant” simply means important.  Suppose you sold a sofa to somebody for

$7 287.  If the buyer wrote you a cheque for  $1 287,  there is an error in the leftmost first digit and you would get  $6 000  less.  That is bad.  If, instead, the buyer wrote you a cheque for  $7 281,  there is an error in the fourth digit and you would get  $6  less.  Bad, but not so bad.  Compare the 7 on the left against the 7 on the right.  Which digit is more significant?  As you can see, the digits on the left are more significant, and decrease in significance as we move towards the right.

The Roles of ‘0’ (Zero)

     Our modern Arabic-based place-value numerical system evolved over many centuries.  It is definitely far superior than Roman numerals.  But what is the symbol ‘0’ for?

     One usage of ‘0’ is to denote nothing, or the absence of something.  In Buddhism, which originated in India, there is this concept of sunyata (from Sanskrit), which means emptiness.  On a stone inscription in a Hindu temple at Gwalior in India, there is a small circle which stands for the digit zero.  Perhaps that was where ‘0’ originated from.  Is life an illusion?  Why then do we use something (a symbol ‘0’) to represent nothing?  Hmmmm .... Anyway, zero as a number is quite convenient to use in everyday life.  If you check up a website of a store for their inventory of jPhone VII mobile phones, and you see ‘0’, you know that they do not have that model of mobile phones in stock.  Without the number zero, they would have to make two separate lists: one for the models in stock and another for those out-of-stock.  By the way, all modern computing, storage and communication depends on 1’s and 0’s in various forms.

     Another usage of ‘0’ is that of a place-holder.  The ancient Chinese had a decimal place-value system for calculations in sand, but they used blank spaces on a grid instead of ‘0’s.  It is difficult to distinguish between say  507 (five hundred and seven)  and  5007 (five thousand and seven), especially if the grid was erased.  As you can see, ‘0’ is used to “hold the forte” in our decimal system.  Without the ‘0’,  507  and  5007  would both collapse to become  57, which means something entirely different.  Similarly, for the decimal  0.0009876,  the leading zeros function as place-holders.  The first significant figure is  9,  the second significant figure is  8,  and so on.  In general, placeholder ‘0’s are needed on the left of the decimal point, as well as to the right of the decimal point. 

Rounding to significant figures

     With the above understanding, here is the easy and sure-fire way to round to a certain number of significant figures:-

Step 1:  Skipping zeros on the left if necessary, identify the first non-zero digit.

Step 2:  From this digit, start highlighting the desired number of significant figures.

Step 3:  Look at the digit just after the last highlighted digit.  If this is 4 or below, do nothing.  If
              it is 5 and above, add one unit to the last highlighted digit.  Carry to the left if
              necessary.

I have highlighted the required significant figures (or digits) in yellow.  After we have identified the first non-zero digit, we just highlight the required number of significant digits, without caring about whether or not anyone of them is zero.  Then we just look at the digit after that to see if we need to round up.

Summary

     To round certain number to  n  significant figures:-

Step 1:  Identify the first non-zero digit from the left. Step 2:  From this digit, start highlighting  n  significant figures. Step 3:  Look at the digit just after the last highlighted digit.  Round up if              this is 5 and above, otherwise do nothing. Always remember to include the place-holder ‘0’s.

Suitable Levels

* Lower Secondary Mathematics (usually secondary 1 ~ equivalent to about grade 7)

* other syllabuses that involve estimation and approximation

[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”.

[#Critique] Is #pi #infinite?

Article / Resource
The infinite life of pi - Reynaldo Lopes

My Comments
This is a well-produced video in which the team took great care in producing an interesting narrative.  The circles were drawn imperfect and shaky, so as to catch the attention of viewers and to highlight them.  A couple of points they missed out:-

1) p being irrational, does not only have an infinite number of digits in its decimal expansion, the digits are also non-repeating.  A rational number (fraction) like 1/7 also has an infinite number of digits, but they repeat: 1/7 = 0.142857 142857 142857 ...

2) near the end of the video, the animators depicted universe < p.  I find the artistic licence disturbing.  I am sure there is some way to artistically depict the known universe having less number of atoms than number of digits of p.

To answer the question in the title of this article "Is p infinite?": No, but her number of decimal digits is.

[Pri20150302MCP] Multiple Choice Questions with Penalty

Question

Solution

The ratio of the number of words spelt correctly to that of words spelt wrongly is 3 : 1.  I use triangle to surround the ratio units.  Multiplying by 10 and 4 respectively, we obtain 30 and 4 units respectively.  The score accrued is 30 – 4 = 26 units which corresponds to 130.  Dividing by 26, we get 1 triangle unit corresponding to 5.  And that is the answer required because it 1 triangle unit represents the number of words spelt wrongly.

Ans: 5

Check: 15´10 – 4´5 = 150 – 20 = 130.   Yes!  Woo hoo!!

Reflection

This question is similar to this fish and chicken cost problem, but there are no monetary units (e.g. dollars or cents) involved.  Also, because of the penalty for wrongly spelt words, we subtract instead of add.  We need to be mindful of the context of the question and to apply the appropriate arithmetic operation.