This problem solving activity has a number focus.
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?
This problem involves students adding and multiplying single digit numbers, and 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.
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?
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.
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 1 | Dog 2 | Dog 3 | Sum | Product |
1 | 2 | 12 | 15 | 24 |
1 | 3 | 11 | 15 | 33 |
1 | 4 | 10 | 15 | 40 |
1 | 5 | 9 | 15 | 45 |
7 x 3 x 4 = 84 and 7 + 3 + 4 = 14
Printed from https://meaningfulmaths.nt.edu.au/mmws/nz/resource/my-dogs at 8:56pm on the 26th February 2024