Multiples of a

Problem

Gill, was 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 was 7a + 9a + 12a + 12a = 40a.

What is the value of your name?

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

Teaching sequence

1. Tell the students Gill’s story and let them find the value of some word, room, say.
2. Make sure that they understand how you find the value of the word. Then ask them to find the values of their names. (That is, just their first names.) Get their partner in their group to check that they have found the right value for their name.
3. Then 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?
4. In groups, the students could then find some way of ending up with a name value of 100a. Can they arrange for both members of their group to have the same value using the same rules? They could be asked to find more than one way to get 100a as their name value.
5. Help the students that need it.
6. The quicker groups can go on to tackle the Extension problem.
7. Get a few groups to report on what they have done.
8. Give all students time to write down something about the answers they got. This should help them to understand what is going on.

Extension to the problem

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

Solution

The answers that you get for the first part of the question will depend upon the names of the students in the class.

Suppose that your name was Garry. The trick here is to put anything you like for G, A and R and then choose Y to make up 100a. There are many ways to do this.

Of course, the same approach will work if you have two different names in your group. The technique here is to 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

Suppose that your name was Gill. Then you could 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. Naturally many other combinations will work too.

N.B. This problem gives lots of scope for exploration. You might like to suggest that the students come up with their own way variation of this problem.