## Can’t see the wood from the tree – an idiom with a mathematical twist

so the saying goes. I brought it up for one reason: the problem we look at today could be formulated in a “wood“ and “tree” context. But, let us start with the original problem.

The problem was given at a final of the AQJM (http://aqjm.fsg.ulaval.ca/), but a similar version was also given at the International Kangaroo Math Contest in Romania a few years back.

The problem from AQJM goes like this:

Consider you have a grid. Starting from the sixth grid-point on the horizontal axis, find the point of minimal vertical coordinate on the vertical axis such that the line between these points cuts more than 10 squares.

The problem given at the Kangaroo contest was: Consider a chess table. Draw
a line on the table such that it cuts as many squares as possible.

One can see that the two problems are not identical, but they are related and bring us to a more general question: given a grid, situated at (0,0) only using non-negative coordinates, and two points, one on each axis, can we tell exactly how many squares will be crossed by the line?

Let’s start with a drawing. Below we have a grid, just 8×8, and we can assume the lower left corner has the coordinates (0,0). We should draw some lines, just to see what kind of situations we will find.

The lines in the figure illustrate different situations: the green one passes just through grid-points, the lilac one cuts some squares and also passes through grid-points, meanwhile
the red ones are just cutting squares. There are no other types of lines, so we can ask: what is defining the type of line?

Getting back to the first problem: we need the point on the vertical axis, with smallest vertical coordinate, such that the line (starting from (6,0) position) will cut more than 10 squares. So, we need to maximize the number of intersected squares.

In problem solving, and not just in mathematics, often the most important task is to decide how to ask a question. We can explain the very same issue in different ways – we can reformulate the information, the question or any other element. And here comes the nice part: you would be asking a question equally relevant for obtaining an answer; however the new formulation will give you new insight into the problem. “Re”-formulation is essential for problem solving and we should teach how to do that or at least
motivate students to try to do it. In the same time reformulation often
requires a flexible knowledge management and that can come only after having a deep
understanding of a subject.

In our particular case, things are simpler. We can “see” that maximizing the number of intersected squares comes down to minimize the number of grid-points the line passes through.

So, now we have another thing to focus on: when will a line go through a grid-point? This question is nice for two reasons.

First, because we need to “interpret” the statement “passes through a grid-point”. What does it mean? Well, it means that we have some integer coordinates for that point of the line, so I can easily identify distances.

Second, because we have integer coordinates on one side and a geometrical illustration on the other hand, we are simultaneously in numbers and geometry. And the geometry will help us to characterize the points our line passes through.

Let’s have a closer look at the green line, in other words triangle AOB. The line passes through D’ of the grid (and E’’), therefore triangles AOB and BDD’, BDD’ and BEE’’ are similar (from Thales). From where we have that the ratio of AO and OB will be the same as DD’ and DB (and, also, E’’E and EB). Since we have integers, once again we reformulate the obtained result: we speak about equivalent fractions.

This point is worth a second of pause, perhaps filled with admiration. From geometry we arrived to equivalent fractions. We know that rational numbers are those that can be expressed as fraction of two integers (denominator can’t be zero), but here and now we have the rational numbers as ratios of distances – which was the way in which Greek mathematicians introduced them in the first place. The problem lead us to see how interconnected numbers and geometry are – and we should understand that considering geometry, algebra, arithmetic as completely separate domains (as often happens in school)
can induce a wrong conception about what mathematics is.

A similar analysis about the lilac line will bring us to the same conclusion.

In conclusion: if we have two points, one with coordinates (a,0) and the second with coordinates (0,b), the line defined by them will pass through grid points if a and b are not relative primes, and the number of grid-points it passes through is equal to the number of common divisors between a and b.

This is an important result, but how many squares will cross the line when a and b are relative primes?

We know that the total number of grid-points inside the rectangle with vertexes (0,0) (a,0) (a,b) (0,b), defined by integers a and b is (a-1)*(b-1). We also know that exactly half of these are on each side of  the line that passes through (a,0) and (0,b) (this rectangle’s diagonal,) that is, (a-1)*(b-1)/2. Given that a and b are relative primes, it means that at least one of them is odd, so the above number is an integer number.

If we focus on one half of the rectangle, that is, one of the square triangles resulting when we cut the rectangle by the diagonal, we can see that the number of squares not intersected by the diagonal is equal to the grid-points inside the rectangle: (a-1)*(b-1)/2.

Therefore, the diagonal cuts all the squares that are not contained inside the square rectangles on both sides of the diagonal:

a*b-(a-1)*(b-1).

If we get back to the AQJM problem: we have a=6 and we are looking for a minimal b such that the number of intersected squares is larger than 10. First, we know that b will be minimal, if  a and b are relative primes (written (a,b)=1). If we were running out of time (there is never too much time on these contests), we can try the numbers that are relative primes with 6 in order, namely: 1, 5, 7, 11, 13, … Hopefully, we’ll find the answer soon enough.

We can also develop the formula we just discovered into a + b – 1. Thus, we know that b must be greater than 5 (for 6 + 5 – 1 = 10). The first prime relative with 6, and greater than 5 is 7, which is the answer.

And, in the end: what about the trees and the wood? Well, just think about those ”planted” woods: trees are planted in grid-points. One could ask: being at one tree on the horizontal edge of the wood, how could I choose an angle to look through the wood so to see only one tree?

What to say? Math is everywhere if you look for it…

## Who drives that car? A follow up on travelling…

Last time we were talking about a beautiful day, cars, constant speed and traveling women. Here is the problem:

Two women from Canada traveled on a beautiful day, by car, along the east-west highway, with the same constant speed. One drove from Town A to Town B, while the other drove from Town B to Town A. Both started at the same hour and crossed each other at certain point on the road. The first arrived at B at 4 in the afternoon while the second arrived at A at 8 in the evening. What is the minimum distance between A and B?

The first thing that would strike me is: how can they travel with the same speed along the same distance, yet one to arrive later than the other?

The only explanation consistent with the data can be that the cities A and B are in different time zones! How can this information help us in finding the minimum distance?

First, we should find out what is the difference in hours between the two towns. Where it is town A? In east or west of Canada? Why? This information is not essential for solving the problem, however it is important since the answer can reflect a certain understanding of time zones and movement of Earth.
If we say that the difference in hours between the two towns is d, then we would have that when one is starting from the eastern city, in the western city the clock shows with d hours less. In other words, the woman from the western city will start traveling with d hours later than the one from the eastern city.
When one of them arrives in the eastern city, the time is with d hours more than in the western city. Therefore, the difference between their hours of arrival comes from the differences in hours at departure and at arrival. Since, the difference between arrival hours is 4, the double of the hour difference. In consequence, we know that there are two hours of difference between city A and B.
How much could be that in km? You can have a look at this site:
in order to learn about time zones in Canada. One reason why I used Canada is that time zones are more or less similar in shape; therefore one could make an estimate. Now, that we know that we have two hours, we could pick up two provinces with 2 hour difference. Then we choose two cities: consider Calgary and Winnipeg. The direct distance between them is more than 1200 km and we could take it as a rough estimate of the minimum distance.
Now, driving more than 1200 km on one day, that’s something else! We met new superheroes from MathLand!

However, it is important to mention that we have some simplifications in the two problems treated in the last two posts. Could you tell?

In the first problem, we supposed that the sun rises at the same time (and at fix hour) in the two cities. Is that true? Well, not exactly and it could be enriching to look for information about geographical place and sunrise hours.

In the second, we supposed the highway goes straight and along the lower part of Canada. Why is that important? From the point of view of the solution, it is not, for the numerical answer it is. In the same time, it takes us to study the variation of the timezones in countries and latitudes. Interesting!

## Heroes of Math Land…

There should be no doubt about it: Elementary Math Problems Land is inhabited by weird, yet fascinating characters!

There are witches who lie about their age (example 1), children with a compulsory need to wrap up every fact in a long story that will give you hard time (example 2), people developing strange rituals to cross the street (example 3)  and also, by people who show extraordinary endurance!  I’ll give you the examples and then, we’ll meet today’s heroes!

Example 1: The bad witch is 4921508 years old. When the Brave Knight asks her age, she takes out four digits from her age such to appear as young as possible. What digits did she take out?

Example 2: Paul says to Pierre: “I have three times the age you had when I had your current age! When you’ll have my age, we’ll have together 112 years!”  How old is Pierre?

Example 3: Mathieu has a habit to cross the street along the line seen in the figure below. What can be x?

As you can see, nobody is ordinary in this magic land! Time to meet some new people!

Two old women started walking at sunrise and each proceeded at a constant speed. One went from Town A to Town B, while the other went from Town B to Town A. They passed each other at noon, and continued on their ways without stopping. The first arrived at A at 4 in the afternoon while the second arrived at B at 9 in the evening. At what time was sunrise on this day?

The nice part in the problem is that, at first, we would not expect the question. The surprise we get with the question is intriguing enough to make us want to solve the problem: is it really possible? Then, if you take the challenge, you actually start thinking about the problem, and it slowly comes together: an image, an idea…ah!

First, we should try to make a drawing. The route followed by each woman is marked with the same color.

Now, it is clear what kind of relations can we write down: at constant speed, the ratio of the overall time on road is the same as the ratio of time on a part of the road (after noon, for example). If we note with t the hour of sunrise we can write the following relations. From A to B, the total time was 12-t+9=21-t, meanwhile form B to A it was “just” 12-t+4, that is 16-t. In the same time, the distance AC is done by the woman going from A to B in 12-t hours and it is the same (as distance) as the distance covered by the second woman, after noon until her arrival, so in 4 hours.

Then we have:

.

From where we have, after subtracting 1 form both sides: .

The simplest way to solve this is to think that 8-t is a positive integer; therefore (16-t) is factor of 20. But, t is less than 8, therefore 16-t is more than 8 and less than 16. That leaves us just the 10 as factor of 20 (why? Can you find other solutions?). If 16-t=10, then t, the hour of sunrise is 6 in the morning. So, one of the women walked 15 hours straight! Isn’t that extraordinary? Told you so…

Let’s change something. This is what I like most. Let’s move something in the problem.

Two women from Canada traveled on a beautiful day, by car, along the east-west highway, with the same constant speed. One drove from Town A to Town B, while the other drove from Town B to Town A. Both started at the same hour and crossed each other at certain point on the road. The first arrived at B at 4 in the afternoon while the second arrived at A at 8 in the evening. What is the minimum distance between A and B?

If you start to dig on this problem, you will see that the new specifications in the problem changed the “nature” of it, i.e. its main mathematical content. But, it can lead to a nice discussion about something else. Let me know how it goes and I’ll come back with the solution in few days.

## Uff, such a long time!

There has been a while since the last post, but many things happened!

First, we are through the semi-finals of AQJM and we are waiting for the results to be released on 21st of April.

Then, quite few of the club participated at the Canadian Math Kangaroo Contest (www.mathkangaroocanada.com ) on March, 27th.

The results from the Virtual Mathematical Marathon (http://www8.umoncton.ca/umcm-mmv/index.php ) were published at the very beginning of the week and we have winners there!  If you missed the winter edition, don’t do the same with the spring one, starting in May!

That is true also for the Saturday Math Club: our spring session starts by the end of April.

Now that we are getting to the end of this intense contest-dominated period, we can relax and concentrate on new problems. One is coming up, pretty soon! Come back and see!

## Semi-finals in the horizon!

Tomorrow, 19th of March 2011, is the semi-final of the competition:

Le Championnat International des Jeux Mathématiques et Logiques

organized in Québec by the Association Québecoise des Jeux Mathématiques (see their site http://aqjm.fsg.ulaval.ca/)

Some of the Saturday Math Club members will be there!

Good luck everyone!

## The “cute” Friday math problem

Do you remember when Mom said that a good, hot chicken soup on cold winter day is the best you can get for your soul? Well, she never asked the chicken about that. So, I took it far from the stomach and I say: there’s no blues a good math problem cannot  fade.

Today, I want to discuss with you a problem given at the semi-finals of the Championnat International des Jeux mathématiques et Logiques. For more information about the contest in Quebec, visit: http://aqjm.fsg.ulaval.ca/

The problem is from the 17th edition of the competition and it was for students from 2nd year of secondary school (in Quebec school system) or 8th graders (in some other provinces of Canada) and up.

What is special about this problem? Well, nothing soooo…special, but it is a nice one. You will never find a commonly agreed definition of what a nice problem is – still, people can agree when they see one.

My pick for this problem was because: how to treat the problem doesn’t come across immediately, so it may seem out of reach for some, but at the same time, once you start thinking about it and reinterpreting the information, the problem unfolds naturally. Besides, the problem lets you see the situation more generally – one can cross the border of particular values and make a larger analysis. Let’s start.

Which are the two, smallest, consecutive integers where the sum of digits of each is divisible by 7?

It might happen that you start to search in the back of your mind for some criteria of divisibility by 7. Searching the web you will find several descriptions for it, but none is at hand and, in fact, you don’t really need that. At least, not before having tried to clearly interpret the information from the problem.

Let’s interpret the information from the problem. If you hear this from your teachers, three times a day, please, don’t blame them, here is the one to blame about this:  http://en.wikipedia.org/wiki/How_to_Solve_It

What if you think what does it mean that they are consecutive numbers – from the perspective of the sum of their digits? Pick up some consecutive numbers and make the sum of their digits. What you should see is that, having two consecutive numbers – in most of cases – the sum of their digits will be consecutive numbers.

So, naturally, comes down to ask when are two consecutive numbers divisible with a digit bigger than 1?

Well, never. The reason is simple: if one of them is divisible by, let’s say, k, then it can be written in the form of a*k, and then the next number is of the form a*k+1, which is not divisible by k.

Now: what does it mean this result for me? The sums should not be consecutive numbers, so when it that possible? It is possible when the last digit of the first number is 9, because then, the next number is rounding up and ends with 0.

How can we write that down? Let’s begin with a two digit number, so we can write a9 (where a is a non-zero digit) and the next will be (a+1)0. The sum of their digits is a+9 and a+1.

We need to have a+9 and a+1 divisible with 7. Immediately we obtain that a has to be 6. But then 6+9=15 is not divisible with 7.

So, we need to make a bigger number. How big should be? The answer is: as big as to have 6+9+9+…+9 divisible with 7, which (taking into account that 9 give 2 as rest from division with 7) is the same as 6+2*k divisible with 7. Which is the smallest k with this property? Yes, you’re right: 4.

In conclusion, the numbers I was looking for are: 69999 and 70000. Nice!

Spice it up! We can ask a more general question by replacing 7 with an arbitrary digit, a, bigger than 1. The reasoning will remain exactly the same until the form of the number: (a-1)9 and a0.

There are two things to see from here right away: first, if a is a digit that divides 9 we are in hot water, (we would need a-1 to be divisible with a, but is not) and, second, some digits are doing better than others. So, we won’t have solutions for 3 and 9. From where, we have a nice midterm problem about proving this inexistence (just an idea!).

For 2, we have immediately: 19 and 20. For 4, we have again: 39 and 40. Five is again a nice digit, because we need to do the second part of the original reasoning by looking for some k such to have 4+k*9 (the same as 4+4*k) divisible with 5. It is immediate that we have 49999 and 50000.

For 6, we need to have 5+3*k divisible with 6. Do we have such a k? (Sounds like hot shower again…) We can rewrite like 3*p+2 needs to be divisible with 6, but then it should be divisible with 3. The villain 2 in the expression ruins it all, but we have another nice midterm exam for primary school: no integer of form 3*p+2 is divisible with 6.

For 8, we have 79 and 80.

Those with powers (of 2) are so lucky! Can you figure out, why? We could ask something else, too. Which are the next two numbers with such properties? (Put it in different way, why had to be the smallest one in the problem?) Give me a feedback on these!

## Welcome!

With this post, we open the Saturday Math Club to everybody out there interested in mathematics.

The Saturday Math Club sessions are held on Saturdays, from 10.30 till 12 for all school levels: from preschool till end of secondary.  The Club also offers geometry classes for those with age 10+ and a special activity for children up to 10: games, puzzles and origami. Classes are held in French.

The club is organized as problem solving sessions with problems grouped around a specific theme. We work mostly with problems of the type given at popular math contests, such as the AQJM contest (http://aqjm.fsg.ulaval.ca/), The International Math Kangaroo contest (www.mathkang.org), American Mathematics Contest (http://www.usamo.org/), etc.

We are ending our winter session on the 16th of April, but spring session starts on the 30th of April, 2011.

We are located at :

Centre Gabrielle-et-Marcel Lapalme

5350, rue Lafond, H1X 2X2