Problem
I have three dogs of different ages. If I add their ages together I get 15. If I multiply their ages together I get 45. How old are my dogs?

What is this problem about?
This problem at its simplest involves the addition and multiplication of single digit numbers. It can also involve the children in considering the factors of a number, in this case 45.

Achievement Objectives
Number (Level 3)
- recall the basic multiplication facts
- write and solve problems that involve whole numbers which require a choice of one or more of the four arithmetic operations

Mathematical Processes
- devise and use problem solving strategies (be systematic, guess and improve)

Resources
3 toy dogs (picture)
Blackline master of the problem (English)
Blackline master of the problem (Maaori)

Specific learning outcomes
The children will be able to:
- Use the properties of multiplication (factors) to solve problems

Teaching sequence

1. Interest the children in the problem – you are in the best position to know what works as a motivation for your class. Using props (pictures of 3 dogs / soft toys) usually works to gain the children's attention.
2. Brainstorm for ways to solve the problem.
3. As the children work on the problem circulate, prompting them to focus on the nature of the numbers they are using.
4. If the children are using guess and check encourage them to think of ways to "improve" with their next guess.
   Why have you selected..?
   What can you tell me about the number 45? Why is that important to this problem?
5. Share solutions.

Extension to the problem
Try a similar problem if the product of the ages is 84 and the sum is 14.
Get the children to write their own problem.

Other contexts for the problem
Costs of chocolate bars

Solution to the problem
We give a sophisticated approach first and then use a tale to guess and improve.

We have to find three numbers that add up to 15. There are lots of them. We also have to find three numbers whose product is 45. There are not so many of them. The clue here is that 45 is not divisible by too many numbers. For instance, the divisors of 45 are 1, 3, 5, 9, 15, 45. What’s more the three dogs have to each have one of these divisors as their age. And what’s even more, none of their ages is bigger than 15 because they all add up to 15!

So we’re now choosing from 1, 3, 5, 9. But the ages can’t be 1, 3, 9 because 1 x 3 x 9 is not equal to 45. So 5 must be in there somewhere. That leaves two numbers to make 9. They must be 1 and 9. So the three ages are 1, 5, 9. Wait on! Does 1 + 5 + 9 = 15? Yes. That’s all right then.

This table shows how our guesses improve until we have the answer.  In this problem we were lucky that one of the dogs was one year old or we  may have needed many more guesses.