This problem helps to develop students' problem solving skills, logical thinking and their ability to work sequentially. It also helps develop rules for mathematical problems. These rules may not be obvious at first.
Students are also challenged to maximise and minimise the problem, by having to find the largest and smallest number (of sheep) that satisfy certain conditions. This is a common theme of mathematics, motivated by the real world and its demands.
This problem builds upon Farmer’s Sheep I, Level 3.
A farmer can see nine sheep if he looks out of any of his four windows. His wife gives him a new sheep. Which paddocks can he put the new sheep in so that he can still see nine sheep from each of the four windows?
What is the largest number of sheep that his wife can give him so that he can still see 9 sheep through each window?
What is the smallest number of sheep that his wife can give him so that he can still see 9 sheep through each window?
If there is to be at least one sheep in each paddock, what numbers can the farmer have and still see 9 sheep through each window?
- Pose the problem.
- Put a large version of the grid up in the classroom and point out the features e.g. where the windows are.
- Explain the rules:
The new sheep must go in a paddock, it cannot be destroyed, hidden, eaten, etc.
The farmer must be able to see a total of nine sheep from the three paddocks he can see from each window.
- Encourage the students to work in pairs and record the problem solving sequences they go through.
- Share solutions. Is there a rule?
Other Contexts for the Problem
The farmer could look at any animal or any crop.
Think of the house as a security camera. The security guard is happy if he can see 9 pieces of art on his surveillance cameras. How many pieces of art can you steal before the security guard is aware that anything is wrong?
The simplest answer is shown above. However, there are 16 answers in all. These are given in Farmer’s Sheep I, Level 3.
To maximise the number of sheep, you want to put the smallest numbers in the corners. There are two reasons for this. The first one builds on the solution to Farmer’s Sheep I, Level 3 but it is not the easiest way. It has the smallest number of sheep go in the corner because
the number of sheep + the numbers in the corner paddocks = 36
As the number of sheep goes up, the number in the corner paddocks goes down to keep the sum constant at 36.
Because you have to have at least one sheep in each paddock, you need to put one sheep in each of the corner paddocks. This gives the answer below as the only possibility. There are 32 sheep.
But it is simpler to look at it in the following way. This way it may be easier to convince the class. Suppose that we put any number in a corner. Reduce this number by one. In order to keep the two ‘window’ sums that include that corner equal to 9, we’ll need to add one to each of the paddocks on either side of this corner. This will change the sheep count by –1 + 1 + 1 = +1. So reducing the number of sheep in the corner paddock increases the total number of sheep by one. This means that we want to put the smallest number of sheep in each corner paddock. This smallest number is 1.
Similarly to minimise the number of sheep you want to put the biggest possible numbers in the corners. (You use the same argument that we used above on the corner paddock. But this time increasing the number of sheep in this paddock will reduce the total number of sheep.) This gives you 20 sheep.
Having got this far you can get any number of sheep from 20 to 32. Start from 20. Then systematically move one sheep from a corner paddock to a neighbouring centre paddock; now add another sheep to the window side that has been reduced by one. This rule should get you from 20 up to 32 sheep.