Letter Designs

In this activity, students base their short cuts and rules on the pattern of coloured counters in the designs.
In question 1, students check Leakhana’s prediction by making the L design with counters. The design should have 6 counters of the same colour in each side and 1 counter of a different colour for the corner. This arrangement will then match Leakhana’s prediction of 2 x 6 + 1 = 13 for the number of counters in the L design.
As the students work out the short cuts for the total number of counters for different L designs in the table in question 1c, they generalise a rule for any L design. It is always challenging for students to write these rules in their own words. One way to write the rule is: find 1 fewer than the number of counters in a side, double the value of this number, then add 1. The students might also try to use algebraic symbols. For
example, if there were x counters in each side of an L design, the total number of counters, y, in the design is given by y = 2 x (x – 1) + 1 or y = 2(x – 1) + 1. Students who manage this should always check that the rule works by trying different values for x. So, when x = 87 (the number of counters in each side), the total number of counters in the design is 2 x (87 – 1) + 1 = 173.
In question 2, the students may find it helpful if they make an L design using the colour pattern shown in the examples. It is more likely that they will then see that an L design with, say, 68 counters in each side will have 68 + 67 counters altogether. This process of generalising is central to the understanding of introductory algebra. So, if there were x counters in each side of an L design, the total number of counters, y, in the design is y = x + (x – 1).
A third way to generalise the L design is developed in question 3. There are 6 counters in each of the 2 sides of the L. While this may suggest that there are 2 x 6 = 12 counters altogether in the design, the corner counter is counted twice. So we subtract a counter to give 2 x 6 – 1 = 11 counters.
This way of visualising the arrangement of counters in these L designs leads to the algebraic expression y = 2 x x – 1 or y = 2x – 1. Here, y is the total number of counters in a design with x counters on each side of the L.
The L design algebraic rules from questions 1–3 are shown in this table:

The rules are, in fact, equivalent.
The rules from questions 1 and 2 simplify to y = 2x – 1 as follows:

In question 4, each T design has 2 green counters and 3 sets of blue counters. For example, the T design with 4 counters in the stem can be visualised in the following way:

We see this as 3 sets of 2 blue counters plus 2 green counters or 3 x 2 + 2 counters altogether.
Similarly, the T design with 5 counters in the stem can be visualised as 3 sets of 3 blue counters and 2 green counters or 3 x 3 + 2 counters altogether.

Visualising this way leads to the generalisation y = 3 x (x – 2) + 2, where x is the number of counters in the stem of the T and y is the total number of counters. This is usually written as y = 3(x– 2) + 2 and is the essence of the rule that the students use to complete the table for question 4b. This algebraic rule simplifies to y = 3x – 4 as follows:
y = 3(x – 2) + 2
= (3 x x – 3 x 2 )+ 2
= 3x – 6 + 2
= 3x – 4
Students who can manage this algebra can confirm their work by completing the calculations shown in this table:

They should see that the two expressions are equivalent and that both give the same answer for any particular values of x.
In question 5, the students see if they can find and explain other rules for T designs. An example of a different T design, with a short cut of 4 + 2 x 2 = 8 counters, is shown in the Answers. For this example, a T design with x counters in the stem has x green counters and 2 sets of x – 2 blue counters. If there are y counters altogether, then y = x + 2(x – 2) or y = 3x – 4, as in question 4 (see above) and question 6 (see below).
In question 6, the students see if they can find their own rules for Cathy’s C designs, using designs made in one or more colours. Some rules and short cuts for 100 counters on the left side are given in the Answers. Those who have shown that they can manage working with symbolic algebra might try to figure out ways to represent their designs algebraically. The designs in the Answers are all shown with 4 counters
on the left side. Three possible solutions for these are shown below.
Solution One
This C design has 3 x 2 (blue) + 2 green counters.

An algebraic rule for this design is y = 3(x – 2) + 2, where x is the number of counters on the left side and  y is the total number of counters used.

Solution Two

This C design has a 4 green + 2 x 2 blue counters.

An algebraic rule for this design is y = x + 2(x - 2).

Solution Three

This C design has 2 green + 2 x 3 blue counters.

An algebraic rule for this design is y = (x - 2) + 2(x - 1).

The three different algebraic rules for the designs above all simplify to y = 3x – 4:

Answers to activity

1. a. The L design below has 2 x 6 + 1 = 13 counters.

b. There are 6 blue counters on each side, so there are 2 x 6 blue counters. There is 1 orange counter in the corner of the L, so there are 2 x 6 + 1 counters altogether.
2. a. A short cut for the first design is 4 + 3 = 7 counters, and a short cut for the second design is 5 + 4 = 9. A rule is: the number of orange counters plus that number minus 1. The L design with 4 counters on each side has 4 orange counters on one side and 3 blue on the other, and the design with 5 counters on each side has
5 orange counters and 4 blue counters. The number of blue counters is always 1 less than the number of orange counters. Or, alternatively, the number of orange counters is always 1 more than the number of blue counters.
b.

3. a. The L design has 6 counters on each side. This suggests that the design has 2 x 6 = 12 counters. But the corner counter is included in each side and has been counted twice, so 1 must be subtracted, giving the total number of counters
as 2 x 6 – 1 = 11.
b. 2 x 256 – 1 = 511 counters
4. a. A short cut for the first design is 2 + 3 x 2 = 8.
The T design with 4 counters in the stem has 2 green counters and 3 sets of 2 blue counters. (The stem includes the middle counter in the top bar.) The T design with 5 counters in the stem has 2 green counters and 3 sets of 3 blue counters. A rule is: 2 green counters plus 3 times the number of blue counters in the stem.
b. Using the rule shown above:

5. a.–b. Answers will vary. Another T design is:

This T design with 4 counters in the stem has 4 green counters and 2 sets of 2 blue counters, which is 4 + 2 x 2 = 8 counters. The same T design with 5 counters in the stem has 5 green and 2 sets of 3 blue counters, which is 5 + 2 x 3 = 11 counters. So a rule is: the number of green counters in the stem plus 2 times the stem minus 2.
c. 746 counters. The rule given in a would give 250 + 2 x 248 = 746.
6. Answers will vary. One rule for the design in  one colour is: the left side plus 2 times the left side minus 2. So, for the first design, this is 4 + 2 x 2 = 8. For 100 in the stem, this is 100 + 2 x 98 = 296. Another rule for this design is: 3 times the left side minus 4. For 100 in the stem, this is 3 x 100 – 4 = 296.
Other rules could be based on designs such as:

For i, a rule is: 3 times the number of blue counters in the stem plus 2. So for 100 counters on the left side, this is 3 x 98 + 2 = 296.
For ii, a rule is: the number of counters on the left side plus 2 times that number minus 2. So for 100 counters on the left side, this is 100 + 2 x 98 = 296. (This is the
same rule as the one for the first one-colour rule above.)
For iii, a rule is: the number of counters in the stem minus 2, plus 2 times the number of counters in the stem minus 1. So for 100 counters on the left side, this is 100 – 2 + 2 x (100 – 1) = 98 + 2 x 99
= 296.