This problem solving activity has an algebra focus.
A farmer can see nine sheep if he looks out of any of his four windows.
His wife buys 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?
In how many ways can the farmer put his 25 sheep so that he can see 9 sheep through each window?
(Every paddock is to have at least one sheep.)
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.
This problem is further developed in Level 4 Algebra problem The Farmer’s Sheep II.
A farmer can see nine sheep if he looks out of any of his four windows. His wife buys 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?
In how many ways can the farmer put his 25 sheep so that he can see 9 sheep through each window? (Every paddock is to have at least one sheep.)
Perhaps the simplest way to do this problem is to notice that you can move one sheep from the top right hand paddock to the left centre one. This keeps 9 sheep in view from each window except the one at the top. If you put one sheep in the centre top paddock this balances up the problem.
There are however many ways of doing this problem but they all need a lot more moving of sheep. Students might suggest other solutions.
The guess and check method will work, but may take some time. When students have a few more solutions, they might begin to see a pattern. Surprisingly, in every case, the numbers of sheep in the four corner paddocks always adds up to 11. Why is this?
The following argument may be a bit complex but some of the more advanced students will be able to see what is happening. Suppose that we added together all of the sheep that can be seen from the four windows. The sum would be 36 (= 4 x 9). But if we add up all those sheep we are adding up the corner paddocks twice and the other paddocks once. So we are adding together every paddock once and the corner paddocks once more. ‘But every paddock once’ gives us all the sheep there are. This is 25. So 25 + the sheep in the corner paddocks = 36. This means that the corner paddocks have to have 11 sheep in them.
A systematic approach gives all the 16 answers below.
Printed from https://meaningfulmaths.nt.edu.au/mmws/nz/resource/farmer-s-sheep at 8:56pm on the 26th February 2024