Gill is playing with her name and with numbers.

If A = a, B = 2a, C = 3a, D = 4a, E = 5a, F = 6a, G = 7a and so on, the value of Gill’s name is 7a + 9a + 12a + 12a = 40a.

What is the value of your name?

Change the rules so that the value of your name is 100a.

Perform calculations with algebraic quantities.

This problem in which students substitute values into their own names, is about powers of 2. It is a precursor to algebra which seeks to generalise number.

Other similar lessons in this series are: __Points__, Level 1, Names and Numbers, Level 2, Make 4.253, Level 3, Go Negative, Level 4 and Doubling Up, Level 5.

### Problem

Gill is playing with her name and with numbers. If A = a, B = 2a, C = 3a, D = 4a, E = 5a, F = 6a, G = 7a and so on, the value of Gill’s name is 7a + 9a + 12a + 12a = 40a.

What is the value of your name?

Change the rules so that the value of your name is 100a.

### Teaching sequence

- Tell the students Gill’s story and let them find the value of their own first names.
- Have partners check that each has found the correct value for their name.
- Ask the students to put themselves into groups all of whose values are the same. Get them to think about their names to see if there is a good reason why they are all in the same group. Is this only possible if they have the same names?
- In groups, have students find ways of making a name value of 100a. Can they arrange for members of their group to have the same value using the same rules? Can they find more than one way to get 100a as their name value.
- As groups work, have students explain their approach and their thinking.
- Notice groups that have interesting and successful strategies, and have them share these with explanations.
- Pose the Extension problem as appropriate.

#### Extension to the problem

Can you get your name to have a value of a + b? How about 3a – 4b?

### Solution

Solutions will depend upon the names of the students in the class.

To make a value of 100a with a name such as Harry: Use any values for H, A and R and then choose Y to make up 100a. There are many ways to do this.

The same approach will work if there are two different names in a group. Give all of the letters that are in common with both names some arbitrary values. Then make up the values for the other letters so that the name values both come to 100a.

#### Solution to the extension

With Gill, for example, let G = a, I = b. This would mean that L would have to equal 0.

For 3a – 4b, let G = 1, I = 2a and L = -2b. Many other combinations will work.

Have your students create their own variations to this problem.