Solve division problems that involve remainders.
Number Framework Stage 7
Pieces of card
Using Materials
Problem: “A postie is sorting letters into pigeonholes at the sorting centre. She has 4 letters to sort into 3 pigeonholes. She is not a very good postie, so she decides to throw the letters into the pigeonholes at random.” Get students to use 4 counters and 3 pieces of card to represent letters and pigeonholes respectively. A mathematician says the postie will put 2 or more letters in someone’s pigeonhole. Explore this claim with counters.
(Answer: It is correct. Trying to avoid putting 2 in any pigeonhole leads to 1 in each. But the fourth letter now has to go in to someone’s pigeonhole.)
Examples: Convince yourself that these claims are true using counters:
For 5 letters and 3 pigeonholes, someone gets 2 letters or more.
For 6 letters and 3 pigeonholes, someone gets 2 letters or more.
For 7 letters and 3 pigeonholes, someone gets 3 letters or more.
Understanding Number Properties:
Find the value of k in each case:
For 23 letters and 3 pigeonholes, someone gets k letters or more. (Answer: k = 8)
For 17 letters and 4 pigeonholes, someone gets k letters or more. (Answer: k = 5)
Using Number Properties
Problem: “Convince yourself that these claims are true without using materials:
For 12 letters and 5 pigeonholes, someone gets 3 letters or more.
For 24 letters and 5 pigeonholes, someone gets 5 letters or more.
For 101 letters and 5 pigeonholes, someone gets 21 letters or more.
Discuss why these all work.”
(Answer: To minimise the number someone must receive, distribute the same number of letters into each pigeonhole. This is done by division. If there is no remainder, then the quotient is the minimum number. However, if there is a remainder, someone must receive another letter, so the answer is the quotient plus 1.)
Examples: Find k in each case:
For 80 letters and 10 pigeonholes, someone gets k letters or more.
For 101 letters and 10 pigeonholes, someone gets k letters or more.
For 45 letters and 8 pigeonholes, someone gets k letters or more.
For 1 001 letters and 4 pigeonholes, someone gets k letters or more.
Problem: “A very mathematically inclined farmer encloses an area ofgrass for his cows by making an equilateral triangle with electric fencingwire. The sides of the triangle are 20 metres long. The farmer places 5cows within the fence. After watching for some time, he noticessomething unusual. No matter how the cows arrange themselves at least 2 of them are within 10 metres of each other at all times. Why is this always correct?”
Hint: Add lines to create 4 equilateral triangles and experiment withplacing cows (5 counters) so no triangle has 2 cows in it.
(Answer: In trying to prevent 2 going in any triangle, you can place 1 in each triangle. But now the fifth cow must go in one of the triangles that already has a cow in it.)
Examples: If the equilateral field has a side of 30 metres draw a diagram and explain why having 10 cows in the field means 2 of them are within 10 metres of each other at all times.
If the equilateral field has a side of 40 metres, draw a diagram and explain why having 17 cows in the field means 2 of them are within 10 metres of each other at all times.
Understanding Number Properties:
The farmer builds an equilateral triangle with sides 10n metres long. How many cows must he put in the enclosed area to ensure there is a pair within 10 metres of each other at all times. (Answer: n2 + 1 cows.)