[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

[OlymPri_20150524MSCR] Round and Round, Twice or Thrice?

Question

Introduction

     If you think the answer is  2 cm (since the wheel rotates two rounds), you are wrong!  This tricky question is taken from a previous year Asia Pacific Mathematical Olympiad for Primary Schools competition.  In the original question, there was no colouring.  I added colours just to make the distinction between the wheel and the track a bit clearer.

Visualisation

     There is some subtlety in this question.  It is something we often do not notice unless we really think hard about it.  To get a handle on what is really happening, let us imagine we made a mark on the wheel in its original position, indicated by a blue dot at a twelve o’clock position.  Let us imagine rolling the wheel clockwise.  Note that the centre of the wheel will move in a bigger circle in an anti-clockwise (American: counter-clockwise) direction within the orange circular track.  Actually it does not matter which way you roll, the centre of the wheel goes round the centre of the track in a direction opposite to that of the turning wheel, which always remains in contact with the track.

Since the blue dot on the wheel turns two rounds, by the time the wheel reaches the bottom half, the blue dot must again be on top or at a twelve o’clock position.  Note however, the point of contact between the wheel and the circular track (indicated by a green dot whenever possible) is not the same as the blue dot!  The green dot is actually a dynamic dot (it is not always the same point on the wheel) whereas the blue dot is always the same dot on the wheel but the wheel is being rotated.

Note that at the halfway point, the green dot is at the bottom of the wheel while the blue dot is on top of the wheel.  Notice also that, relative to the wheel, the green dot goes in the opposite direction as the blue dot and they actually crossed over somewhere along the way!  Actually the green dot has made one-and-a-half turns with respect to the wheel.  Remember that the green dot is not a fixed point on the wheel, but it measures how much the wheel and the track have been in contact.

As the wheel continues to roll back up, the green dot makes another  1½  rounds.  So altogether the green dot moves through  3  rounds while the blue dot rotates only  2  rounds around the centre of the wheel!  The green dot is the one that matters.

Solution

     circumference of track (measured by green dot) : circumference of wheel = 3 : 1

     Since radius is proportional to circumference,  the radius of the track is  3  cm.

Remarks

     Yes, the solution is that short.  But the thinking behind it is profound.  But do we need to draw all the diagrams as in the visualisation above?  I did that to explain to you.  Actually I imagined it in my mind.  If imagination is difficult, you can act it out by drawing a big circle and then using a small coin to simulate the rotation around the track.  I actually drew a rough sketch by hand  to convince myself that my thinking was accurate.

     Notice that I used a proportionality argument.  If you know how to use the concept of proportion, you can make the working short and sweet.  There is nothing wrong in using the formula                     circumference = 2p ´ radius.

This formula just says that the circumference of a circle is proportional to its radius, and the constant of proportionality is  2p.  Your working would look like this

                                  circumference of track = 3 ´ circumference of wheel

                                      2p  ´ radius of track = 3 ´ 2p  ´ radius of wheel

After cancelling out the  2p,  you would get

                                       radius of track = 3 ´ radius of wheel = 3 cm

You get the same conclusion, but using the proportionality method, you do not need to bother about the  2p.

     By the way, the centre of wheel traces out the path of a circle of radius  2 cm.  See the red arrow in the first diagram on top.  The path traced out by the blue dot looks like an  epicycle.  In the old days,people thought that the sun, moon and planets rotated around the earth in epicycles.  This also reminds me of Spirograph,which is a toy that allows you to use your coloured pencils to create very beautiful patterns.  [Click to search for images of Spirograph and patterns produced.]

Heuristics Used

H01. Act it out

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

Suitable Levels
* Primary School Mathematics Olympiad
* anybody game for a challenge relating to imagination, circumference and lengths

[S2_20150513IXS] The Power of Repeated Chunks

Question

Introduction

     This is a question on indices (taught in secondary 2 or 3) that looks daunting.  There is a way to unravel it.  Look at the equation carefully.  Can you see anything that looks like it is repeated?  What about that small little “+1”?  How do you deal with it?

Reminders and Tips

    

Solution

Summary

     The key to solving this question is to apply the laws of indices and recognising repeated chunks.  This allows one to see the overall structure of the expression and this often leads to a simplification of the expression involved.  It is good to recognise the powers of small prime numbers.  Like atoms, prime numbers are the building blocks of all whole numbers.  Using prime bases allows us to compare indices more easily.

Heuristics Used

H04. Look for pattern(s)   e.g. repeated chunks

H09. Restate the problem in another way

H10. Simplify the problem

H11. Solve part of the problem

H13* Use Equation / write a Mathematical Sentence

Suitable Levels

* Lower Secondary Mathematics (usually Sec 2)

* “Elementary” Mathematics or “E Maths”

* Additional Mathematics (revision)

* other syllabuses that involve indices

[Pri20150306MOD] A Telescope Too Far?

Question

Introduction

     This is a nasty multiple-choice question to be set for a primary 6 (~ grade 6) pupil for a test.  It could also be a time trap as the pupil might spend a lot of time to no avail just to try to score that miserable mark.  With advanced knowledge, we can solve this using telescoping sums (a.k.a. the method of differences), where many terms cancel and the sum can be shortened, much like a telescope.

     However, this solution looks like over-kill.  Is there a solution that is more accesible to a primary 6 pupil?  Read on!

A Simpler Solution?

     We observe that a denominator of  5  is common.  In fact, since every term has an even factor, we have a common factor of 2 in the denominator also.  That means every term in the sum can be expressed as ¹/₁₀ of something [ Heuristic H09 ].  We can quickly work out the first few partial sums and express each of them as  ¹/₁₀ of something.  [ Heuristic H04 ]

     Tabulating the results [ H02 & H03 ], we observe that each partial sum is  ¹/₁₀ of something slightly less than ½.  So we try to express them as ½ minus something.  How?  We can subtract to find out.  [ H05 ]  For example,  if  ⁵/₁₂ = ¹/₂ – ???   then   ??? = ¹/₂ – ⁵/₁₂ = ¹/₁₂.

     Later on, we notice   [ H04 again ] another pattern: the factors in the denominators of the subtracted quantity matches the last two denominators of the last term.  If we considered the full sum, then what would those factors be?  (See the part highlighted in yellow).  These would be 20 and 21, as per the last term of the series.  Using this pattern, we work out the required sum as follows

That is the answer!

Heuristics Used

H02. Use a diagram / model

H03. Make a systematic list

H04. Look for pattern(s)

H05. Work backwards
H09. Restate the problem in another way