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My dogs

Student Activity: 

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?

Achievement Objectives:

Achievement Objective: NA3-1: Use a range of additive and simple multiplicative strategies with whole numbers, fractions, decimals, and percentages.
AO elaboration and other teaching resources
Achievement Objective: NA3-2: Know basic multiplication and division facts.
AO elaboration and other teaching resources

Specific Learning Outcomes: 
Use the properties of multiplication (factors) to solve problems
Devise and use problem solving strategies (be systematic, guess and improve)
Description of mathematics: 

This problem involves students in adding and multiplying single digit numbers, and in considering the factors of a number. By using a systematic approach, students can show if there is more than one possible solution to the problem.

Activity: 

The 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?

Teaching sequence

  1. Interest the students in the problem and confirm that it is well understood.
  2. Have students suggest ways to begin the problem.
  3. As the students work on the problem, prompt them to focus on the nature of the numbers they are using.
  4. If the students 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

Pose a similar problem in which the product of the ages is 84 and the sum is 14.

Challenge the students to write their own problem.

Other contexts for the problem

Costs of food items, ages of family members

Solution

A range of number combinations have a sum of 15.  The problem also requires three of these numbers to have a product of 45. There are fewer of these because 45 is only divisible by 1, 3, 5, 9, 15, 45. The three dogs must each have one of these divisors as their age. Together their ages must add to 15.  

Therefore the choice is from 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 one of the ages must be 5. That leaves two numbers to make 9. They must be 1 and 9. So the three ages are 1, 5, 9 which do have a sum of 15.

This table shows how guesses can be improved until a solution is found. Many more guesses may have been needed if one of the dogs is not one year old.

Dog 1Dog 2Dog 3SumProduct
12121524
13111533
14101540
1591545

Solution to the Extension

7 x 3 x 4 = 84 and 7 + 3 + 4 = 14

AttachmentSize
MyDogs.pdf308.78 KB
AkuKuri.pdf355.27 KB