The Clumsy Tiler S
Use a pattern to count in a geometrical situation.
Devise and use problem solving strategies to explore situations mathematically (make a drawing, use equipment).
This problem is one of three parallel problems. Each one takes up the basic theme of Fred and his difficulty with tiles. But this one has a statistics flavour while the other two have geometric and numerical aspects (see The Clumsy Tiler G and The Clumsy Tiler N, respectively). All of the problems are at Level 4.
One of the points of these lessons is to show that by changing a problem by just a little, it ends up in another strand of the curriculum.
The current lesson deals with the different ways that the tiles could be laid. The student have to keep track of them as they are laid. They especially have to be careful to keep track of where the broken tile was about to be put. In the process they should be able to find a pattern.
Like The Clumsy Tiler N, the numbers that we are dealing with here form a geometric sequence. Such a sequence requires a fixed number to multiply each term in order to get the next one. The sequence 2, 6, 18, 54, … is a geometric sequence where the fixed number that does the multiplying is 3. Fred’s problem is about finding how many ways that the 1 square metre part of Sue’s floor can be tiled.
But the geometric arrangements here are important too as they give some insight into the way that shapes fit together. The geometrical side of the problem leads on to tessellations of the plane and to visualising objects in 2- and 3-dimensions. Hence it helps to lay the foundations of a feeling for geometry that bear fruits later in Cartesian geometry and in trigonometry.
Fred was a man who laid tiles. He had this beautiful big 1 metre by 1 metre square tile for part of Sue’s new kitchen floor in front of her fridge. But he broke it as he was putting it down.
Fred couldn’t replace the tile but he did find four square tiles to fill the space of the big one. He laid three of these tiles perfectly but as he was putting the last one in place he broke it.
Fred couldn’t replace these new smaller tiles, so, in the hole where the broken one should have gone, he laid four even smaller tiles. He laid three of these tiles perfectly but as he was putting the last one in place he broke it.
This happened another two times. How many possible ways could Fred have laid the tiles on Sue’s kitchen floor?
How many possible ways could Fred have laid the floor if he had broken 100 tiles?
- Ask the students if they have been in any rooms that have tiles on the floor. (It’s likely that they will have been in toilets where there were square tiles.)
How do they lay tiles like that?
- Tell them about Fred.
What did his first tile look like?
What did the second one look like?
How were his first three tiles laid? In how many ways could this be done?
Help the students to realise that, because there is a fridge next to the tiles, the square area is fixed at one end. This means that there are four ways to tile the floor with the first three tiles.
- Repeat Fred’s story. Get them to work in groups to solve the problem.
- Give them a hand if they need it. Let the groups that finish quickly try the Extension problem.
- Ask one or two of the groups to tell how they solved the problem. Make sure that everyone has a good idea of how to do it.
- Some of the students who worked on the Extension problem might like to report back too. If no one has solved this check that everyone understands what the problem is. You might like to suggest that they ask their parents for help and leave the solution for another day.
- Give the students time to write down a solution in their books.
How many different ways are there of tiling Sue’s square metre, if Fred replaces one tile at each stage by 9. (Of course, he breaks one of these at each stage too.)
When Fred is tiling with the first three tiles he can leave a space in 4 places. Hence at the first stage, there are 4 ways that the floor can be tiled. We show these 4 ways below
Now at the next stage, the three new tiles can go into the blank space in the diagrams above. Again, in each diagram they can be laid in 4 ways. This gives rise to 16 tilings.
At every new stage, each diagram leads to 4 new tilings. So we progressively get 16 x 4 = 64 tilings, then 64 x 4 = 256 tilings, and then 256 x 4 = 1024. Since for the first part of the question there are five stages, we get 1024 tilings at this point.
Now what if there were 100 breakages? Spot the pattern. Each new stage gives rise to a factor of 4 more tilings. So for 100 stages we’ll get 4 x 4 … x 4 tilings (with 100 4s).
The only difference with 9 tiles is that there are 9 possible ways that we could have a tile missing. At the first stage then, there will be 9 possible ways to tile the square. But at the next stage we get 9 different possibilities for each of the 9 tilings. So after 5 stages we have 9 x 9 x 9 x 9 x 9 = 59049.
After 100 stages we’ll have 9 multiplied by itself 100 times. That’s a colossal number. It’s about 26 with 94 zeros after it. Something like
260 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000.