Fred lays tiles. He has a big 1 metre by 1 metre square tile for Sue’s new kitchen floor.

But he breaks it as he is putting it down.

Fred can’t replace the tile but he finds four square tiles to fill the space of the big one.

He lays three of these tiles perfectly but as he is putting the last one in place he breaks it.

Fred can’t replace these new smaller tiles, so, in the hole where the broken one should have gone, he lays four even smaller tiles.

He lays three of these tiles perfectly but as he is putting the last one in place he breaks it.

This happens another two times.

What was the size of the last tile that Fred breaks?

How big would the last whole tile be if Fred had broken 10 tiles?

This problem explores the pattern of diminishing lengths (of tiles), as these halve and halve again. Students are challenged to find a particular value in this pattern.

The numbers 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. Such sequences and series and their generalisations are an important part of mathematics.

There are two parallel Level 4 problems in this series: The Clumsy Tiler G and The Clumsy Tiler S

Some square pieces of cardboard

### The Problem

Fred lays tiles. He has a big 1 metre by 1 metre square tile for Sue’s new kitchen floor. But he breaks it as he is putting it down.

Fred can’t replace the tile but he finds four square tiles to fill the space of the big one. He lays three of these tiles perfectly but as he is putting the last one in place he breaks it.

Fred can’t replace these new smaller tiles, so, in the hole where the broken one should have gone, he lays four even smaller tiles. He lays three of these tiles perfectly but as he is putting the last one in place he breaks it. This happens another two times.

What was the size of the last tile that Fred breaks?

How big would the last whole tile be if Fred had broken 10 tiles?

### Teaching sequence

- Discuss floor tiles and have students identify places they know are tiled. Ask:
*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? - Pose the problem and have them in groups or pairs to solve it, keeping a clear record of their process.
- Have groups that finish quickly work on the Extension problem.
- Ask one or two of the groups to tell how they solved the problem. Provide an opportunity for others to question if they don't understand.
- Some of the students who worked on the Extension problem might like to report back too.

#### Extension to the problem

How big would the last whole tile be if Fred had broken 100 tiles?

### Solution

Fred broke 5 tiles. Each time he broke a tile he replaced it with a tile that was half as long as the previous one. We know this because each time four of the new tiles replaced the one that was broken.

Fred began 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 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

#### Extension:

If Fred 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?