[OlymUSec_20150412BDP] Guessing Cheryl’s Birthday

Question

Introduction

     This “Primary 5 mathematics” (actually an upper secondary Olympiad) logic puzzle has gone viral.  It has been making its rounds in various forums in Singapore and overseas, stumping adults and children alike.  It is actually a parody of an old puzzle.  Can it even be solved?  It seems that there is no information given by each parties that we can exploit.  Actually there is!  In a subtle way ...

Solution

     In the beginning, everybody knows that Albert knows only the month and Bernard knows only the numerical day of the month.

     When Albert tells us “I don’t know when Cheryl’s birthday is, but I know that Bernard does not know too.” he is leaking out information (from his knowledge of the month) that the day of the month appears more than once and cannot be (June 18 or May 19).  Actually, the original phrasing is more like “If I don’t know when Cheryl’s birthday is, then Bernard does not know too.”.  The person who set this question merely changed the names of the people and the dates, without appreciating the subtle but crucial difference between a statement of fact and an implication (an “if ... then ... ” statement). 

     Ruling out June 18 and May 19, we also know that Albert knows that the birthday month is neither June nor May.  Otherwise, how would he have been so confident in saying that he knows Bernard would not know Cheryl’s exact birthday?  So we can eliminate those months.

     Bernard acknowledges the above state of affairs and the embedded hint.  With the choice narrowed down and with his knowledge of the numerical date, he now knows Cheryl’s birthday.  Since we know that Bernard knows Cheryl’s birthday, we know that it cannot be a numerical date that appears more than once (otherwise he would not have been able to know).  So we can cross out July 14 and August 14.

     Now Albert would telepathically thank Bernard for this helpful hint.  Because now he is able to deduce Cheryl’s birthday with his knowledge of the month.  That would mean that this cannot be a month with two candidate dates.  We blot out the August dates and see for ourselves the only remaining possibility.

Conclusion: Cheryl’s birthday is  July 16.

Remarks

     This puzzle was solved using the process of elimination and analysing our knowledge of what each party knows and can know.  Thus we successively narrow down the possibilities until the answer becomes obvious.  Here we learn that

     knowledge of other people’s knowledge can itself give us knowledge. 

This principle is actually employed in cryptology (the use of secret codes) which finds applications in fields like banking, the military (cf. interesting story of how the German Enigma code was broken in WorldWar II) and communications.  As an example, radio communication can tell the enemy of troop positions and warn of an impending attack, and that is why radio silence is imployed as a precaution.  Sensitive information in certain organisations is restricted on a “need to know” basis.

Suitable Levels
* Upper Secondary Olympiad
* other syllabuses that involve knowledge or epistemology
* application of mathematical principles in real life
* for all people interested in logic puzzles

[EM_20150412SCA] Is your Airport Design career “taking off”?

Question

An Airbus 380 has constant acceleration of  1 m/s2.  Its takeoff velocity is 280 km/h.  How long must the runway be at a minimum to allow the plane to take off?

Introduction

     A practical question for airport design, perhaps?  I present two solutions.  The first uses a graphical method (speed-time graph) which is in the (“Elementary”) Mathematics syllabus and the other uses formulas for motion under constant acceleration taught in Physics.  Whichever method is used, remember to convert from km/h to m/s.  The target velocity is  ⁷⁰⁰/₉ m/s,  and the time to achieve this is  ⁷⁰⁰/₉ s  starting from rest,  since the acceleration is  1 m/s².

Solution 1 [“Elementary” Mathematics, speed-time graph]

     The speed-time graph is very useful because it is able show the acceleration (as the gradient or slope of a straight line) and at the same time the area under the graph gives the numerical value of distance travelled.  If the acceleration is constant, we usually we get a trapezium.  But since the aeroplane starts from rest, we get a triangle (see diagram below).  All we need to do is to calculate the area under the graph and get the answer.
     It is interesting to observe that if we used the average speed  ⁷⁰⁰/₁₈ m/s,  we would also get the answer because the area under the graph (yellowish green rectangle) is the same as the area of the triangle.  This trick works for constant acceleration, but it may not work in other situations.

Solution 2 [Physics, constant acceleration]

     We use the important formulas  v = u + at  and  s = ut + ½at².  In our example,  u = 0  because the initial velocity is zero (the airplane starts from rest).  This makes our calculations very easy.  If we compare the two methods, you find that the calculations are very similar, and we get the same answer.  Remember that speed = |velocity|  the magnitude of velocity.  In this relatively easy problem, the velocity means the same as speed because we are going in a straight line and in one direction only.  In other situations, this may not be so.

Heuristics Used

H02. Use a diagram / model

H05. Work backwards

H09. Restate the problem in another way

H13* Use Equation / write a Mathematical Sentence

Suitable Levels
* GCE ‘O’ Level “Elementary” Mathematics
* GCE ‘O’ Level Physics
* other syllabuses that acceleration, speed and distance
* precocious kids who always want to learn more

[OlympiadGRN20150406] Grid Ratio Ninja

Perhaps you are an expert at Fruit Ninja, slashing fruits mercilessly with precision using your much-feared finger.  Now can you do the mathematical equivalent below?

Question

Solution

     We would be getting nowhere fast if we took a random approach.  We are only allowed two slashes.  Let us look at the question for more clues.  We have a  5  by  6  grid, giving a total of  30  cells.  The total number of parts in the ratio is  1 + 2 + 3 + 4 = 10.  10  parts correspond to  30 cells,  so each part is  3  cells.  So the number of cells in each portion are  3,  6,  9  and  12.

     Note that if we pair up  3  with  12  and  6  with  9,  we have  15 cells each.  We can just slash the grid down the middle to achieve this.

     We now need ratios of  1 : 4  and  2 : 3  for the remaining parts.  We certainly can achieve this if we were allowed two more slashes and if we used horizontal lines, like this:

However, we only have one more slash.  Can we make this slash do the work of the two horizonal slashes?  Yes!  Replace the two horizontal slashes with a single slanting slash that passes through the mid-points of the horizontal slashes.  This gives four trapezia.  The reason why it works is because for each portion, an extra triangular area is added but it is offset by the removal of a triangle of the same area, like this:-

We can check that the ratios of the areas of the portions are as required.  Done!

Heuristics Used

H02. Use a diagram / model

H04. Look for pattern(s)

H05. Work backwards

H08. Make suppositions

H09. Restate the problem in another way

H11. Solve part of the problem

[OlympiadXPH20150405] Seats in a Hall as Pigeonholes?

Question

There are  25  rows of seats in a hall, each row having  30  seats.  If there are  680 people seated in the hall, at least how many rows have an equal number of people each?

Introduction

     This is a mathematics olympiad type of question for primary / elementary school, I believe.  Let us make sure we understand the question.  Each row of seats has a certain number of people, which I shall call the “headcount”.  We want all the headcounts to be as different as possible, and yet we do not want so many rows to have the same headcount.  The number of rows with the same headcount should be kept as low as possible.  How low is low?

Solution 1

     One strategy to solve this is to start filling up the empty seats with as many as possible.  Indeed this is the approach taken by the official “model answer”, which for copyright reasons I cannot show.  However I am going to do something even better: to explain the solution visually.  Recall that the sum of an arithmetic progression is 

                   ½ ´ number of terms ´ (first term + last term)

     After filling up the empty seats with different numbers of people, we would have filled up  30 + 29 + ... + 7 + 6 = ½ ´ 25 ´ (30 + 6) = 450  seats.  This is stage 1, shown in orangey-yellow in the diagram below.  We have now  230  people remaining to be seated. 

     For stage 2, we try to fill in the remaining seats with as many people as possible but keeping the headcounts all different.  So we ramp up  6 to 30,  7 to 29,  8 to 28, ... etc.  The additional number filled up is  24 + 22 + ... + 2 = ½ ´ 12 ´ (24 + 2) = 156.  This is stage 2, shown in green.  We have  74  seats remaining.

     For stage 3, we ramp up  18 to 30,  19 to 29,  20 to 28, ... etc.  The additional number filled up is  12 + 10 + ... + 2 = ½ ´ 6 ´ (12 + 2) = 42.  This is stage 3, shown in blue.  We have  32  seats remaining.

     For stage 4, we ramp up  19 to 30,  20 to 29,  21 to 28, ... etc.  The additional number filled up is  11 + 9 + 7 + 5 = 32.  We are done.  This is the final stage, shown in pink. 

Discussion

     It turns out that there is another way to look at the problem.  The above solution can be depicted in a “ball and bin diagram” as shown below.

There are  25  balls in the diagram, each representing a row’s headcount.  There are  4  rows with 30 people,  4 rows with 29 people, ...,  3 rows with 26 people,  3 rows with 25 people,  2 rows with 24 people, and  1 rows with 26 people.  We can actually calculate the totals for rows with repeated headcounts.  For example in the diagram, there are  4  layers of balls with each layer representing 30 + 29 + 28 + 27 (shown in dark turquoise) and the sum is = 4 ´ [½ ´ 4 ´ (30 + 27) ] = 456.  Those with rows with three of the same headcounts total up to  153  (shown in pink).  The rows with two of the same headcounts total up to  48  (shown in yellowish green).  There is one and only one row with  23  (shown in dirty green).  All this give a grand total of  680.  We can calculate the grand total  for every ball and bin diagram in this fashion.  We want a grand total of  680 and there must be 25 balls.  Using this diagram and a generalised version of the Pigeonhole Principle, we can have a very short and sweet solution.  Before that, let me briefly explain the Pigeonhole Principle.

Pigeonhole Principle

     Let us say there are 30 pigeonholes and 31 envelopes.  We can represent this with a ball and bin diagram using 30 columns (for pigeonholes) and 31 balls (representing the envelopes).  If you try to put one envelope into each pigeonhole it is impossible.  One of the pigeonholes will have two envelopes.  In the ball and bin diagram, there will be at least column with two balls.  If you try to put everything flat to one layer, that is 30 balls and you still have one ball left.  You will have to put this remaining ball somewhere on the second layer.  This illustrates the Pigeonhole Principle.

     Likewise if you have  91  balls and  30 columns, there must be one column with  4  balls.  If it is all  3  balls, that fills with only 90 balls.  Your remaining ball has to go somewhere on the fourth layer.  This illustrates the Extended (or Generalised) Pigeonhole Principle.

Solution 2

     I am going to further extend the Pigeonhole Principle to Pigeonholes with Valuations (with some number attached to each configuration e.g. grand total).  We have 25 balls and 30 bins (columns) and we want to fill up the grand total number as quickly as possible achieving a grand total of  680, using as few layers as possible.

     Starting from  30  downwards and, using  3  layers,  we have  8  balls for each layer, filling all the colums for  30,  29,  ... ,  23.  We now have used up  3 ´ 8 = 24 balls that represents a grand total of  3 ´ [½ ´ 8 ´ (30 + 23) ] = 636.  There are 44 seats left over, but only one remaining ball.  The best we can do is to put the remaining ball at  22  for one layer.  That leaves  22 unallocated seats.  So  3  layers are not enough.  We need  4  layers.

     To show that  4  layers are enough, we can imagine filling up  4  layers of  30 + 29 + ... + 25.  That gives a grand total of  4 ´ [½ ´ 6 ´ (30 + 25) ] = 660,  with  20 left over.  This can be handled by putting the remaining ball on  20  for one layer.  And we are done.  The latter configuration is an alternative configuration to the one given in the model answer.  But most importantly, it works.  We have shown that at there must be least  4   rows with the same number of people each.  

Note

1.  There can be more than one set of  4  rows with the same number of people each, but we just need to show that there exists (one set or more of)  4   rows with the same headcounts.

2.  One can imagine configurations that have  5  rows with the same number each, but it would not be “least”.

3.  To answer the question in the title of this article, the seats are not the pigeonholes.  Rather, the pigeonholes (or bins, or columns) represent possible numbers of people in a row.  Each ball in bin  j  represents a row that has  j  people.

Heuristics Used

H02. Use a diagram / model

H03. Make a systematic list

H04. Look for pattern(s)

H05. Work backwards

H06. Use before-after concept

H07. Use guess and check

H08. Make suppositions

H09. Restate the problem in another way

H11. Solve part of the problem

[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”.

[Pri20150308CSR] Charitable Savings Ratios?

Question

Introduction

     Here is another one of those Singapore Mathematics problems that are two-variable simultaneous equations in disguise.  The key to solving this question quickly is to exploit the fact that the amount of donations are the same in this case.

Solution

     First, read the question and translate the information into a diagram [H02. Use a diagram / model].  I use different shapes (e.g. circle and square) to envelop the different types of units.

     Since we have ‘-80’  for both Sharon and Ryan [H04. Look for pattern(s)], we may deduce that 1 ‘circle’ unit  (5 minus 4 ‘circle’ units) is equal to 3 ‘square’ units (10 minus 7 ‘square’ units).  That means 5 ‘circle’ units is 15 ‘square units’.  [ H10. Simplify the problem]

     By comparison again, we realise that 5 square units is 80 [H05. Work backwards] and hence 15 ‘square’ units is 240.

Ans: Sharon’s savings was $240 at first.

Check

Before    $240   $192      5 : 4

After      $160   $112     10 : 7

Solution makes sense.

Heuristics Used

H02. Use a diagram / model

H04. Look for pattern(s)

H05. Work backwards

H10. Simplify the problem

[Pri20150306APC] Apples-with-Pears Comparison

Question

Introduction

     This question involves money and looks rather challenging, because there are a few unknown quantities.  To solve it, we use the concept of unit costs, so that we can make an “apples-to-apples” ... er ... I mean “pears-to-apples” J comparison of the prices.

Solution

     Since absolute dollar amounts are given, we can quickly solve [H11. Solve part of the problem] for the total costs of pears and apples as follows [H02. Use a diagram / model]:-

     Now we know that the total cost (in dollars) of pears is 45 and for the apples it is 50 (5 more than 45).  Although we do not know the absolute numbers of pears and apples, we know their ratio.  So let us write these down as, say, ‘square’ units. 

     Dividing the total costs by the numbers gives the unit costs, which we know only in ratio terms.  So let us use, say, ‘circle’ units to denote these.  But we know that the unit cost (in $) of an apple is 0.50 less than that of a pear.  And that is equivalent to 5 ‘circle’ units.  From here [H05. Work backwards], we quickly work out the cost of a pear (15 ‘circle’ units) as $1.50.

     Ta da!

Heuristics Used

H02. Use a diagram / model

H05. Work backwards

H11. Solve part of the problem

[Pri20150306PPP] A Perspective for Perplexing Parallelograms

Question

Comprehension

     Let us first be clear about what is given in the problem.  I have indicated the numerical values of the areas in the diagram below.

Planning

     First we construct  HK  parallel to  AB  passing through  E.  [ Heuristics H02 & H09 ]  This makes the problem easier to solve because there are congruent triangles and there are connecting ratios along the sides of the parallelogram as well as along the diagonal.  Namely  BF : FA = BE : ED = BK : KC.  We denote the  Area of DBKE by  x, and observe that it is the same as  Area of DBFE  by congruency (you can shift and rotate  DBKE  to get  DBFE).   Area of DAHE = Area of DAEF = 1 cm²  because of congruent triangles.  [ H04 ]

     Let us try to figure out  x [ H11 ].  Making further observations [ H04 ], note that  Area of DADB = Area of DCDB (big congruent triangles) and Area of DHDE = Area of DGDE  (congruent triangles).  See diagram below.

     Because of all these congruent triangles, the parallelogram AHEF and EGCK are forced to have the same areas (area measures), even though they may not be congruent:-

      Area EGCK = Area DCDB – Area DGDE – Area DBKE

                           = Area DADB – Area DHDE – Area DBFE 

                           = Area AHEF = 2 cm².

We conclude that  Area DBKE = Area DBFE = ¹³/₅ – 2 = ³/₅  [cm²].  [ H05 ]

Here is the important thing:-

If triangles/parallelograms have the same height, their area ratio equals their base ratio.

     [ This is different for similar triangles/parallelogram. Do not confuse these two situations. ]

     Area DBFE : Area DAEF = BF : FA = BE : ED = BK : KC  = 3 : 5.

Now we can quickly fill in the areas of the remaining pieces as follows:-

      Area ABCD = 2 ´ Area DABD = 2 ´ (³/₅ + 2 + ⁵/₃) = 8 ⁸/₁₅  [cm²]   Done!

Heuristics Used

H02. Use a diagram / model

H04. Look for pattern(s)

H05. Work backwards

H09. Restate the problem in another way

H11. Solve part of the problem

Thinking Back

     After solving a problem (especially a difficult one), it is always good to think back and recollect what we have learnt by solving the problem.  We have used the following important geometrical facts

#1   If two shapes are congruent (that means you can shift, rotate and/or reflect so        that they coincide), then their areas are equal. #2   If triangles/parallelograms have the same height, their areas’ ratio equals their        bases’ ratio. #3   Sometimes, two shapes can have the same area even if they are not congruent.

      Drawing a construction line splits the diagram into various parallelograms and triangles with the same height, for which we can compare the ratios.  Using congruent triangles, we realise that two of the smaller parallelograms have the same area even though they are not congruent.  With the right perspective, we can deal with these perplexing parallelograms.  Using comparison, we found the area of the small triangular piece  x, and from there we use fact #2 above to work out the rest.

[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

[Pri20150306TTT] The Tricky Triangle

Question

Introduction

     This is another one of those tricky primary school mathematics questions involving areas.  A perfunctory glance at the area seems to suggest there are four pieces.  Later you might realise that you can think of it as two quarter-circles with two little 45°-45°-90° isosceles triangles removed. 

Plan

     Our plan will be to first find the areas of the two quarter circles and then to subtract the areas of the isosceles triangles.  This is our usual divide-and-conquer strategy [ Heuristics H10 & H11 ].  Notice that the two quarter-circles can be rearranged [H09] into a semi-circle with radius 10 cm.  Simple enough.

What about the two exised triangles?  Notice that the longest sides  (the sloping sides) of the triangles (highlighted in green) are each equal to the radius  10 cm  of the quarter-circles, simply because they, by touching the arcs, are themselves also radii of the quarter-circles.

However, the problem seems to be that we do not know the base and the height of each triangle.  Examiners for Singapore Primary School mathematics like to set this sort of questions involving areas of isosceles right-angled triangles, in which you are given only the length of the hypotenuse (the longest side).  How to tackle this kind of situation?  By using our imagination!

Imagine that the two triangles are brought together.  This forms a larger right-angled isosceles triangle.  However, now you realise it is half of a 10 cm by 10 cm square.  You can also imagine turning the triangle around until one of the 10 cm sides is horizontal.  Treating this as the base, the height of the triangle is 10 cm.  Either way, you are able to solve it and get the same answer.

     All that is left now is to subtract this from your area of the semi-circle found earlier.

Solution

   Shaded Area [in cm²]

= Area of two quarter-circles – area of two triangles

= Area of semi-circle – area of combined triangle

= ½ ´ p ´ (10) ²  – ½ ´ 10 ´ 10

= 50p  – 50

Ans: Shaded area = 107.08 cm².

Heuristics Used

H09. Restate the problem in another way

H10. Simplify the problem

H11. Solve part of the problem

Commentary

     We solved this problem by breaking it down into smaller problems.  Since areas are unchanged when you shift them, or turn them, or reflect them, we are able to arrange the two quarter-circle pieces into one semi-circle.  We can also combine the two triangles into a larger triangle for which we know the base and the height.  By breaking down the problem and transmuting these smaller problems into equivalent problems, our task becomes much simpler, allowing us to get the solution quickly.

Please refer to this similar problem.