Wednesday 18 September 2019

Giving iPhone the iFinger

Our mobile phones were getting creaky, so with our new cellular contract my wife and I each updated to an iPhone 8. Previously, I had an iPhone 6. My wife had a fairly old Blackberry.

The transition has not been smooth.

One particularly distressing problem was that at one point our daughter's phone (an iPhone 6S) stopped receiving messages from our new phones. We live in Edmonton and my daughter lives in Montreal, and what made this problem more dreadful was that our iPhones reported that the messages had been delivered. Even more annoying, our daughter's phone was able to receive messages from her sister (who has an iPhone 8 and who lives in Edmonton).

My wife discovered this problem, and she was extremely upset about it––so much so that she seriously considered resurrecting her old Blackberry and adding it to our cellular account just so she could reliably exchange texts with our Montreal daughter.

Prior to this we had visited the Genius Bar at the Apple Store for an unrelated problem, and they had altered some of our iPhone settings. So, thinking that the messaging problem might be due to some wonky settings, I had a long chat with Apple Support. This did not fix the messaging problem and Apple Support suggested it might be caused by our cellular provider. (Seems unlikely, since both our daughters have different providers than we do.)

Following this my wife and I spent a lot of time tinkering with our phone settings, and, using only our own iPhones, we were able to replicate the problem.

We found that if my phone had the iMessage setting toggled OFF, and if my wife's phone had the same setting toggled ON, my phone did not receive messages from my wife's phone.

But what really stunned us was that when the phones in question just swapped settings, so that my iMessage was ON and my wife's was OFF, the problem went away.

So, basically, the problem seemed to boil down to this:

Some iPhones are not able to receive messages from other iPhones.

This made no sense whatsoever.

More details

We think we have partially isolated the cause of this weirdness, and that it has nothing to do with our service providers. It appears that it might be a flaw in the Apple messaging system that has to do with how the iPhone may or may not be linked to other Apple devices.

I have 3 other Apple devices, my wife has an iPad Mini.  My iPhone is linked (or synced) to my other Apple devices. My wife's iPhone is not linked to her other Apple device,

We had to learn a bit about Apple's message system, and you you may want to skip this section if you are familiar with iPhones:

An iPhone deals with two different types of messages: iMessages and text messages.

iMessages have speed and size advantages over text messages, but only an iPhone can send and receive iMessages, and the phone can do this only if the iMessage setting is toggled ON. Any other mobile phone, including an iPhone with iMessage toggled OFF, is limited to text messages. (You can toggle the iMessage setting by going to Settings > Messages > iMessage.)

So what happens when you try to send an iMessage to, say, an Android phone?

No problem! 

The Apple message system is designed so that if an iMessage is sent to any phone that does not have iMessage capabilities, the iMessage is automatically converted to a text message. The people at Apple Support confirmed that this is supposed to be the case.

This should mean that if iPhone #1 sends an iMessage to iPhone #2 that happens to have iMessage toggled OFF, then the iMessage should be converted to a text message which can then be received by iPhone #2.

As I mentioned, when my phone had iMessage OFF and my wife's had iMessage ON, my iPhone did not receive messages, and my wife's iPhone reported that her messages were delivered. In this situation we discovered, quite by accident, that my wife's messages were actually delivered to all of my linked Apple devices, but no messages, text or otherwise, were delivered to my iPhone.

(In case you are wondering what happened when my wife's phone also had iMessage turned OFF,  Then, regardless of my phone's iMessage setting, ON or OFF, my iPhone, along with all of my other devices, always received the message. And her phone also received any message that I sent to her.)

I checked with my daughter in Montreal to see what her iPhone settings were when we were having problems messaging her. In order to avoid data overages she had toggled iMessage OFF, and she confirmed that her iPhone was linked to another Apple device.

What's this about the iFinger?

There is at least one other setting, the Cellular Data setting (Settings > Cellular > Cellular Data), that appears to interact with the message system.

In the problematical case, where my wife's iPhone had iMessage ON and, and my phone had iMessage OFF, the situation sometimes changed if my phone had Cellular Data toggled OFF. In this case, when my wife tried to send me an iMessage, it sometimes went into the blue iMessage bubble. If she held her finger on the bubble, her iPhone prompted her to send the message as text. When she accepted, the message was indeed sent as text, and my iPhone received it.

However, the iFinger procedure was not always available. More often than not, her iMessage was immediately sent. She had no opportunity to touch the blue bubble and we were back to the case where my iPhone did not receive a message while her phone reported as delivered.

The solution to all this, which was suggested by our Edmonton daughter, is to configure the iPhone 8 as follows:

  • turn iMessage OFF,
  • keep Cellular Data ON,
  • stop linking your iPhone to other devices.

This is how our Edmonton daughter has set up her iPhone, which explains why our Montreal daughter was able to receive messages from her.

Wednesday 16 May 2018

Dividing by a fraction

This post was prompted by conversations with April Brown and Leann Miller, teachers from the Peace Wapiti School Division in Alberta whom I met at our April 2018 SNAP Math Fair Workshop at BIRS.

Invert and multiply

Is there any real-life circumstance where the invert-and-multiply rule is naturally applied? Here are two instances when I used this rule without being aware that I was using it.

The rule I am talking about is this one:

where b and c are numbers and [A] is pretty well anything.

Red Deer to Edmonton

I'm driving from Red Deer to Edmonton. That's 144 km. I know from past experience that my average speed will be 90 km per hour. How many minutes will it take me to get to Edmonton?

Did you solve it like this:
144 divided by 90 is 1.6, so it takes 1.6 hours, which is one hour and 36 minutes, which is 96 minutes.
That's how I got the answer. Putting it in mathematical notation:

I could have obtained the answer by first converting the speed from km per hour to km per minute –– 90 km per hour is 90 km per 60 minutes which is 1.5 km per minute –– which gives this solution:
To travel 144 km at 1.5 km per minute will take 144 divided by 1.5, or 96 minutes.
In mathematical notation:

The two solutions are equally valid, which means that

So, dividing by 90/60 is the same as inverting the fraction and multiplying by 60/90.


In this next problem, the numbers work out nicely. In the actual task, the numbers were not quite so friendly.

Building a shelf

You have been asked to construct a 28 inch wide shelf using 2 × 4 lumber.

The shelf will be 8 feet long (which is convenient because two-by-fours come in that length). The shelf will have the boards snug against each other, that is, no spaces are allowed between the boards. 

How many boards will you need? 

Since 28 ÷ 4 is 7, we will require 7 boards. But there's a catch: two-by-fours are not 2 inches by 4 inches. The two-by-fours that you buy at a lumber store are milled a quarter of an inch on all sides, so that their actual dimensions are 1½ inches by 3½ inches.

So the question becomes:
How many  3½ inch wide boards are needed to make a width of 28 inches? In other words, what's 28 divided by 3½ ?

If it were me, I would shamelessly bring up the calculator on my smart phone and punch in 28 ÷ 3.5 to get the answer: eight boards.

To avoid the awkward ½ in a mental calculation, you might note that two 3½ inch wide boards are together the same as one 7 inch wide board.  Since four 7 inch boards will span 28 inches exactly, this means that eight 3½ inch boards will do the job.

If you've followed this, we've just inverted-and-multiplied again. Using the fact that 3½ can be written as 7/2 , this is what we did (after a slight rearrangement of the notation):

The expression on the left is 28 divided by 3½ while the expression on the right is 28 multiplied by the reciprocal of 3½.

Some additional comments

I learned the invert-and-multiply rule in grade 8. At that time, some of us had difficulty deciding what number we should use to divide or multiply by.  In the Red Deer to Edmonton problem, should one multiply by 60/90 or by 90/60?  Should the final calculation be

To overcome difficulties in situations like this, our physics teacher gave this useful advice (which today is sometimes offered on the web):

Always check that the units cancel properly!

When the units are included, the confusion about 60/90 versus 90/60 becomes:

Which of the following should we use

or, equivalently, which of these should we use

Here's what the physics teacher meant by the units cancelling properly:

In the expression on the left, the km units cancel, leaving us with an answer in minutes –– which is what we want. The units in the expression on the right don't cancel and so we are left with km2/minute, which doesn't make sense in this problem. 


Finally, it is interesting to note that there is a connection between the invert-and-multiply rule and the rule for the distribution of the subtraction sign.

In general, without the usual fractional notation, the invert-and-multiply rule looks like this

a ÷ (b ÷ c) = a ÷ b・c.

Here's the rule for the distribution of the subtraction sign 

a – (b – c) = a – b + c.

In both rules, the sign outside the parentheses flips the signs of the operations inside the parentheses.

There are really a lot of math connections happening here. There is a tie-in with the BEDMAS convention for the order of precedence of operations. There is the potential lead-in to additive and multiplicative inverses. There are even hints about how to remove the mystery from rules like a-negative-times-a-negative-is-a-positive. 

Tuesday 20 February 2018

Nine Men in a Trench

Nine Men in a Trench is a math puzzle invented in 1917 by H. E. Dudeney. Historically, it was a dark time. The battles of World War One were never ending, and casualties were horrendous. At the Battle of the Somme, which lasted over four months, the combined casualties for the Allies and the Germans totalled an unimaginable 1,200,000 people.

WWI battles featured trench warfare, which must have given rise to the setting for Dudeney's puzzle.

Photograph from Canadian government archives showing Canadian soldiers at the Somme. Those recesses dug into the sides of the trench are called funk holes, and they are a factor in Dudeney's Nine Men in a Trench puzzle. 

In contrast to all this darkness, we have this wonderful photo of a student from Contentnea-Savannah School in North Carolina. She is displaying her puzzle at a math fair.

Photo posted with permission.

This is the puzzle:

Nine Men in a Trench

Move the red marker from the right end to the left end. You can move markers into the spaces, but cannot double them up. Nor can you jump a marker over another.

(If you look closely at the photo, you can see markers on the puzzle board behind the student.)

The puzzle can be found on our math fair website (link provided below). It is a simplification of Dudeney's puzzle. Here is his original version, complete with funk holes and soldiers:

Nine Men in a Trench

Here are nine men in a trench. Number 1 is the sergeant, who wishes to place himself at the other end of the line––at point 1––all the other men returning to their proper places as at present. There is no room to pass in the trench, and for a  man to attempt to climb over another would be a dangerous exposure. But it is not difficult with those three recesses, each of which will hold a man.

How is it to be done with the fewest possible moves? A man may go any distance that is possible in a move. 

So where's the math??

The Nine Men in a Trench puzzle does not explicitly involve arithmetic. At SNAP, we promote the use of math based puzzles in the classroom, but puzzles like this sometimes prompt the above question.

The truth is that much of mathematics does not involve arithmetic.  Nine Men in a Trench is a sorting problem, and sorting problems are dealt with in both mathematics and computing science courses.

Like all sorting problems, Nine Men deals with a list of objects that have to be put into a specific order, with restrictions on what manoeuvres are allowed when reordering the objects. 

To solve the Nine Men puzzle, students have to figure out procedures to use that will allow them accomplish the sort. Although they are dealing with a 100 year old puzzle, when students work with this puzzle they are truly doing up-to-date mathematics. Beyond memorizing and practicing the usual algorithms of their math courses, when they work on this puzzle they are also beginning to invent algorithms of their own making.

A more challenging adaptation of the puzzle makes it very clear that it is a sorting problem:

Five Men in a Trench

Put the tiles in numerical order from left to right. Stay within the playing board. You can move tiles into the spaces, but cannot double them up. Nor can you jump a tile over another.

We can ask for the minimum numbers of moves required to solve this version (although that should probably not be the point of the puzzle for children). I have been able to do it in twenty-five, but I don't know if that is the smallest possible number. As in Dudeney's puzzle, a move is defined as sliding a single tile any number of positions, following the rules for moving a tile of course.

More about connections between math and puzzles

Teresa Sutherland and I have put together a small booklet about the connections between school mathematics and the puzzles on our math fair website. It includes pointers to both the Alberta Math curriculum and the American Common Core State Standards. Teresa is a Designer/Director/Teacher from Maryland for the STEAM-based Odyssey of the Mind Inspired Spontaneous Challenge Program. 

You can obtain the book from the SNAP math fair website. Or contact me, Teresa, or Sean Graves, respectively at 




The original Dudeney puzzle (and others) can be found in either of the following books:

Friday 16 February 2018

Who needs algebra anyway?

These five balance and weighing puzzles are intended to be solved using only manipulatives. Some of them are suitable for a puzzle-based math fair. They are fun to solve and require a lot of mathematical thinking.

No algebra is needed, but if you have algebra at your disposal, it is interesting to see how each puzzle can be modelled as a system of linear equations.

There are three different types of boxes. All boxes with the same colour weigh the same.

These boxes balance.

How many blue boxes are needed to balance one red box?

So, the solution for puzzle #1 is not difficult: The balance at the top left says that two yellow boxes are the same weight as three blue boxes. So, attending to the balance at the top right, replace the two yellow boxes with three blue boxes to get

By removing three blue boxes from each pan (which  maintains the balance) we see that five blue boxes will balance one red box.

Here's one that is a little more interesting:

There are three different types of boxes. All boxes with the same colour weigh the same.

These boxes balance.

How many blue boxes are needed to balance one red box?

And one that is more challenging:

There are three different types of boxes. All boxes with the same colour weigh the same.

These boxes balance.

How many blue boxes are needed to balance one red box?

Or this one:

There are three different types of boxes. All boxes with the same colour weigh the same.

 These boxes balance.

How many blue boxes are needed to balance one red box?

Here is one that is about inequalities.

There are three different types of boxes. All boxes with the same colour weigh the same.

These boxes balance.

These boxes will not balance.
Which side of the balance will tilt down?

Puzzle #5 is from Sam Loyd, where it was not given as a balance and weighing puzzle, but was set up as a "Tug o' war" problem:
In a tug of war contest, four stout boys could tug just as strong as five plump sisters. Two plump sisters and one stout boy could hold their own against two slim twins. If three plump sisters and the slim twins team up against one plump sister and four stout boys, which side will win?


There are major differences between using manipulatives and using algebra to solve these particular puzzles.

1. There are no negative numbers when using manipulatives. In the following situation, you cannot isolate the red box by removing the yellow box from the left pan because there is no balancing yellow box that can be removed from the other pan.

Whereas with algebra, we readily subtract y from both sides of the equation

r + y = 5b 
                         to get
r = 5b – y.

2. There are no fractions when using manipulatives. In puzzle #4, I did not ask the question
How many blue boxes are needed to balance one yellow box? 
because there is no answering it using manipulatives alone.

It could be deduced that two and one-half blue boxes are needed to balance one yellow box, but you cannot get this via manipulatives unless your manipulatives are sliced into smaller pieces beforehand. (And you have to use at least a minimal amount of algebra to determine the size of the slices.)

If we use algebra to solve the puzzle, we fairly quickly obtain the fact that

 y =  52 b,

even 'though you can argue that this is of dubious meaning in the context of the physical puzzle.

A more complete answer to the question in the post title:

Sometimes you really don't need algebra, but algebra can make thinking a lot easier.

Friday 12 January 2018

Real world math ??

Here's something that percolates through K-9 math education:

Always relate your projects, exercises, and tests to real-life and the real-worldTeach students that math is necessary in the real-world. Students want to know how math applies to their world. They want to see its practical value.  Doing this will make it appealing to them. 

Well, duh! Looks like an obvious way to make math more appealing. But things "gang aft agley".

Consider this scenario: You are making up a question for elementary students that requires multidigit addition.  So you come up with:

Ageing Relatives

(Also known as "When in doubt, add!")

John’s aunt is 35, Tom’s uncle is 42. What is their combined age?

Without the trigger word "combined" why would you even think of adding the ages?

Young children try to make sense of new experiences. Unfortunately, there is not much opportunity for sense-making with this problem.

The setting (two people of different ages) is real, but the context (the necessity for doing the math) is missing. To solve the Ageing Relatives problem, children are obliged to do math in a setting that does not provide a rationale for actually doing the math. Give them too many problems like this one, and they may start operating according to this principle:

When there are two numbers and it is not clear how to use them, 
the proper thing to do is to add them.

There is evidence that this happens. A while back, this problem swept through the web:

Robert Kaplinsky gave the question to 32 grade eight students, and 24 of his students came up with a numerical answer.  (You can see his discussion here along with reactions from other teachers.) It is not clear whether the students provided numerical answers because they wanted to please the teacher, or whether they acted reflexively and just did some arithmetic with the numbers that happened to be available.

Here's an algebraic variant that my daughter sent me:

Here is a more advanced version of the "math without rationale" syndrome.

Training a Mouse

(A magic function problem)

Alex has been training a mouse to find a reward by carrying out a certain task. He wants to know how long it will take the mouse the find the reward on its fifth try.

The time that it takes the mouse to find the reward is modelled by

T(n) = 0.04n2 – 6n + 30,

where T is the time in seconds and n is the number of times the mouse has previously tried the task.

How long will it take the mouse to find the reward on its fifth try?

A magic function is one with a unfathomable connection to the setting, but which nevertheless must be used to obtain a solution for the problem. Since the derivation of the magic function is unknowable, it carries with it this message:

Any useful parts of mathematics will forever remain beyond our understanding.

In truth, the sole intent of a magic function problem is to have the student do something with the given function –– in this particular example, it is to evaluate the function when n = 4.  In problems like this, the function may or may not be realistic.

I don't know if magic function problems can be fixed without scrapping the function, in which case one should discard the setting and leave the question as a procedural exercise.

Below are a few more vexatious real-world problems. I tried to show how to fix the problem by providing alternate settings that still retain the math task while avoiding the awkwardness of the original setting. Some of the fixes employ a setting that is not real-life. (I believe that most students will accept a pretend-world context as long as it does not attempt to pass itself off as their world.)

The Diner Menu

(Math is everywhere?)

Andrew saw the following sign at a diner. If he bought one of each item and spent $7.50, how much did the drink cost?

Burger3x +  0.05
Drinkx + 0.10

The question is nice and short. But have you ever seen product prices listed like this?

Here's a possible fix that uses a fictitious setting.

(I'll agree with you that the fix is not much better than the original question, but at least it removes that incredibly bogus real-world setting.)

On the forest moon of the planet Endor there lived an Ewok family with 3 children. The parents had given their children some Baubles and Bangles. Here's what each child has:

Infant Ewok1 Bauble
Pre-school Ewok1 Bauble & 2 Bangles
Teen Ewok3 Baubles & 1 Bangle

The children want to exchange their Baubles for Bangles, and the parents have agreed to do this.

Unfortunately, none of the Ewoks could remember how many Bangles a Bauble was worth. All they could remember was that, taken all together, the totality of the children's Baubles and Bangles had the same value as 18 Bangles.

After the exchange, how many Bangles will each Ewok child have?

The following is a different fix, also not real-world, and it's a bit too wordy, but the setting does provide a plausible reason for the way the data is presented.

Mrs Boychuk and her two young daughters were visiting an historical tower that had three observation levels.  Leading between the levels were one or more red staircases sometimes followed by extra wooden steps.

All of the red staircases were identical. Mrs Boychuk asked Freddy to count the number of steps in a red staircase, and asked Frankie to count the number of wooden steps between each level.

Freddy dashed as fast as she could to the top level and enthusiastically reported that there were 90 steps in all. In her exuberance, she didn't count the number of steps in a red staircase.

This is the data they collected:

LevelsRed stairsWooden stairs
Level 0 to Level 11 staircase0
Level 1 to Level 21 staircase10
Level 2 to Level 33 staircases5

Of course, Mrs Boychuk was a bit peeved, since she was going to ask the youngsters to add in columns and find the total number of steps. But then she realized that she had a new problem to ask:

How many steps are there between each observation level?

Cookie Boxes

("The cart before the horse it is!")

Before her coffee break, Ms Kowalchuk had prepared 24 boxes of cookies, and tied each box with a red ribbon. When she came back from her coffee break she noticed that the ribbons were removed from 4 boxes and 7 boxes were missing.

How many boxes with red ribbons were left?

Many of us have had a lot of experience with word problems. When I see one like this, I tend to react reflexively and proceed to solve it without considering the reasonableness of the problem.

How did Ms Kowalchuk know that 7 boxes were missing without first counting how many boxes were left and subtracting that from 24? In other words, the answer to the question had to be known before there was enough information to ask the question.

We can make the math task seem more reasonable by changing things so that the setting itself explains how Ms Kowalchuk noticed that 7 boxes were missing without already knowing the answer she is seeking.

Before her coffee break, Ms Kowalchuk had prepared treats for her community league book club. She put out 24 plates each with a cupcake and two cookies. When she came back, she saw immediately that four plates only had cookies on them, and she counted seven plates that were completely empty.

How many plates were left that still had a cupcake?

Apple Inventory

(Using pliers to hammer a nail?)

A grocery store manager noted that on Tuesday, for every 3 McIntosh apples the store sold, they sold 4 Red Delicious apples. On Tuesday they sold 24 McIntosh apples.

How many Red Delicious apples did they sell?

My friend is an excellent handyman. He vents when he sees someone using the wrong tools. For me, using ratio and proportion to track inventory also calls for a vent. It is a situation where an inappropriate mathematical tool is being used to do a real-world task for which other tools are available and would certainly be used.

As in the previous problem, a change in the setting will fix things and still retain the desired math.

A recipe for making 3 quiches calls for 4 eggs. A pastry shop wants to prepare 24 quiches using that recipe.

How many eggs will they need to make 24 quiches?

The Cargo Truck

(Blind spots.)

Use the following information to answer the question below

A total of 10 packages are arranged in the back of a cargo truck as shown in the diagram below. One large package has the same mass as two medium packages. One medium package has the same mass as three small packages.


How many small packages need to be loaded onto the right side of the truck to balance the load?
A.    8
B.    9
C.   12
D.   13

I had more fun with this problem than any of the others because it illustrates one of the hazards of imposing a real-world setting. The setting has blind spots –– properties that are extraneous to the math task, but which provide alternate and more reasonable approaches to solving the actual real world problem. And when this happens it can turn a good math task into an annoying one.

For the Cargo Truck problem, the real-world setting is a cargo truck loaded with packages of differing masses. The real-world problem is to balance the load. The solution method is imposed by whoever set the problem: add more small boxes to the load.

In real-life, the most obvious way to fix an imbalance would be to reposition the packages already in the truck. (And if that were permitted, then there is an answer where no extra packages are needed.)

Also, a real-life cargo company would know the masses of the individual packages rather than their masses relative to each other. Knowing the individual masses would lead to a different approach to the problem. (Also, this raises the "cart before the horse" objection: it is not clear how the company would obtain the relative masses without first knowing the actual masses of each package.)

The diagram itself could be misleading for some students. It suggests that there is not enough space for the extra small packages that are required as ballast.  (Thirteen are required unless you reposition some packages, but we have already discussed that).

On the other hand, the imposed solution method involves some interesting math, and it worth finding a setting, real-world or not, that works.

Here are two different fixes. This first one retains the notion of balance:

There are three different types of boxes. All green boxes weigh the same,  all blue boxes weigh the same, and all orange boxes weigh the same.

These scales balance. These scales balance.

How many more orange boxes are needed on the right pan to make the scales balance? 

(You cannot move the boxes on the left.)

The second modification drops the demand that the setting must be 100% real-world. This setting uses the concept of money, so it should be familiar to the students, yet different enough to signal that the question is not trying to pass itself off as truly real-world.

A country uses three different coloured coins. Green ones are called buckazoids. Blue ones are called halfzoids. Orange ones are called mini-zoids.

Each buckazoid is worth two halfzoids.
Each halfzoid is worth three minizoids.

Mary and John each have the money shown:

Mary's  money. John's money.

Their mother has promised to give John enough minizoids so that he has the same amount of money as Mary.

How many minizoids should their mother give to John?

Am I against using real-world contexts? No. but, but sometimes they can produce really bizarre questions. Paradoxically, extensive use of real-world settings can cause a dissociation between math and the real world.

For more on this, see this post by Nat Banting and this Globe and Mail column written by Sunil Singh a few years ago.

Problem Sources

The Diner Menu is due to Cathy Yenca,  who posted the question in the sequence of articles by Dan Meyer about pseudocontext in math problems. There are many more examples in those articles.

The Cargo Truck is a question from the sample grade 6 Alberta PAT.  Alberta Education considers this question to be illustrative of the more challenging ones in the PAT.

The other problems are rewordings, paraphrases, or amalgams of questions that I found on worksheets and other resources posted on the web.

Thursday 14 December 2017

Math fair workshop at BIRS

Every spring for the past 15 years there has been a 2-day math fair workshop at the Banff International Research Station. In 2018, this wonderful event runs from the evening of April 27 until noon April 29.

The workshop is all about how to run a math fair that emphasizes puzzle solving with lots of  interaction between the math fair visitors and the student presenters. The workshop participants will primarily be teachers, including some who have organized highly successful math fairs in their schools. More details about the workshop can be found here. I hope you will consider coming.

Some of the participants at the April 2017  workshop

The workshop sessions are held in the TransCanada Pipelines Pavilion. It's a superb venue, with the all the capabilities and facilities you would expect from a leading research institution. 

TransCanada Pipelines Pavilion. Photo Courtesy of The Banff Centre.


Here are some examples of the sort of puzzles that we will be working with. These happen to involve arithmetic operations, but that is not obligatory –– all that is required is that the puzzles be mathematically based. 

Fill in the digits

(a) (b)


(a) Put the digits 1, 2, 3, and 4 into the squares to make a correct sum. Use all four digits.

(b) Put the digits 1, 3, 6, and 8 into the squares to make a correct sum. Use all four digits.

(c) Put the digits 1 through 7 into the squares to make a correct sum. Use all seven digits.

The following two examples are both based on the same idea and show how puzzles can be adapted to challenge students at different levels.

Crosses and Sums

In the cross above, the numbers from 9 through 12 have been placed in the squares in such a way that horizontal and vertical sums are the same.

( 10 + 8 + 11 =  9 + 8 + 12. ) 

In each of the crosses below, the squares have to be filled with all of the digits from 1 through 5.

(a) (b) (c)

In each of (a), (b), and (c) put the remaining digits from 1, 2, 3, 4, 5 into the empty squares to make the horizontal and vertical sums the same.


In the following figure, the digits from 1 through 7 have been placed in the circles so that the sums along the lines are the same:

1 + 7 + 6 = 2 + 7 + 5 = 4 + 7 + 3.

In each of the following, the circles have to be filled with the digits from 1 through 7. In each case, three of the circles have already been filled.


(a) Place the numbers 1, 2, 3, and 7 in the circles so that the sums along the lines are the same.

(b) Place the numbers 4, 5, 6, and 7 in the circles so that the sums along the lines are the same.

Both the Crosses and Sums and the Spokes puzzles are variations of the Spoke Sums puzzle from our math fair website, which is in turn a simpler version of a much older puzzle Henry Ernest Dudeney and/or Boris A. Kordemsky. I don't know who originated the puzzle –– during that era, puzzlers frequently "borrowed" from each other without giving credit.

For more examples of math fair puzzles, visit our SNAP math fair website.

If you are planning to come to this year's workshop and have some favourite math-based puzzles or games, please bring them and share how you have used them in your teaching.