The Clumsy Tiler N
Handle smaller and smaller fractions.
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 number flavour while the other two have geometric and statistical aspects (see The Clumsy Tiler G and The Clumsy Tiler S, 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 lengths of tiles. The students have to keep track of them as they get smaller and smaller or, more precisely, as they halve and halve and halve again. In the process they should be able to find a pattern.
Now, in fact, 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 a large term in this sequence. This is followed in the extension by considering what happens when the terms are added. Here we are looking at a geometric series. Such sequences and series and their generalisations are an important part of mathematics and come up later in the high school curriculum as well as at university.
Fred was a man who laid tiles. He had this beautiful big 1 metre by 1 metre square tile for Sue’s new kitchen floor. 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. What was the size of the last tile that Fred broke?
How big would the last whole tile be if Fred had broken 10 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 big was it?
- Repeat Fred’s story. Get them to work in groups or pairs 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.
Extension to the problem
How big would the last whole tile be if Fred had broken 100 tiles?
It’s probably best to work this one through carefully. From the question he broke 5 tiles. Each time he broke a tile he replaced it with a tile that was half as big as the previous one. We know this because each time four of the new tiles replaced the one that was broken.
So if Fred started off with a tile whose side length was 1 metre, after the first break he was using tiles whose side lengths were 1/2 metre. The second break gives tiles with side length 1/4 metre; the third break 1/8 metre; the fourth break 1/16 metre and the fifth break 1/32 metre.
If Fred really got into this breaking groove and broke tiles 10 times then what fraction of a metre would the length of his tiles be? Let’s go back to the first case with 5 breaks. The fractions there can be re-written as 1/2, 1/(2 x 2), 1/(2 x 2 x 2), 1/(2 x 2 x 2 x 2) and 1/(2 x 2 x 2 x 2 x 2). In each case there is one extra 2 for each extra break. So for 10 breaks we would get 10 2s. So the fraction would be 1/(2 x 2 x 2... 2 x 2 x 2), where there are 10 2s inside the bracket. This gives us 1/1024
If Fred really got into this breaking groove and broke tiles 100 times then what fraction of a metre would the length of his tiles be? Let’s go back to the first case with 5 breaks. The fractions there can be re-written as 1/2, 1/(2 x 2), 1/(2 x 2 x 2), 1/(2 x 2 x 2 x 2) and 1/(2 x 2 x 2 x 2 x 2). In each case there is one extra 2 for each extra break. So for 100 breaks we would get 100 2s. So the fraction would be 1/(2 x 2 x 2... 2 x 2 x 2), where there are 100 2s inside the bracket.
How big is 2 x 2 x 2... 2 x 2 x 2 where there are 100 2s multiplied together? How big is 1/(2 x 2 x 2 ... 2 x 2 x 2), where there are 100 2s in the denominator?