Displaying Postcards
Apply algebra to the solution of a problem
Devise and use problem solving strategies to explore situations mathematically (be systematic, use a table, use algebra).
Interpretation of the words is important in this problem as it is in many word problems. There is a lot of information that has to be gleaned from the problem – more than just equations involving multiples of 4 and 5. For instance, the fact that Paul can’t have a prime number of cards is fundamental to the solution.
The pattern in this question is generated from the rule given in words and the students may not need to translate the problem into algebra in order to solve it. In solution Option 1 we show this and complete the problem largely by way of a table.
If you do take an algebraic approach it might be useful to use letters that help you to remember what they represent. So we suggest s, j, p rather than a, b, c, because s links to Sally, j to Jane and p to Paul. This is worth considering in any algebraic problem.
If you go down the algebraic route you can speed things up by realising that you are dealing with a Diophantine equation. These are equations that only have whole number (or integer) solutions. Such equations often contain additional information that you can use to solve them. For instance, in this problem we have 3 equations and 4 unknowns. However, we can solve the equations because the solutions have to be whole numbers. (See also Pigs, Goats and Sheep and Rings and Diamonds.)
One important Diophantine equation that was solved in 1975 and that was unsolved for over 300 years, is the subject of Fermat’s Last Theorem. This asks if it is possible to solve xn + yn = zn for integers x, y, z, where n is a whole number bigger than 2. The problem was originally posed by Fermat. It took a tour de force by the mathematician Andrew Wiles before the problem was solved.
Problem
Sally, Jane and their brother Paul collect postcards. They each buy a big project book to display them in. The books weren’t necessarily all the same size.
The eldest sister Sally, who is always organising everyone, suggested that they should put the same number of postcards on each page to make the books look neater and that they should put more than one post card on each page.
Each of the three children put their postcards in a pile on the floor.
Sally said, "If I place my postcards four to a page, I'll have three cards left over".
Jane said, "If I had just one more postcard I could group mine five to a page".
The youngest child, Paul said excitedly "I can fit mine perfectly on each page already and I guess I've got the most and I'll be the first to get to100 anyway".
They decided to count their postcards and see who had the most. To Paul's disappointment they discovered that they all had the same number.
How many postcards do they each have? And how many pages do their project books have?
Teaching sequence
- Interest the students in the problem by talking about patterns in the multiples of 9. List the multiples of 9 (9, 18, 27, 36, 45, 54, 63 ) on the board and ask:
Do you know what these are? What is the next number?
What patterns can you see?
Does anyone know an easy way to remember the 9 times table? - If no-one suggests it show the students the "trick"for the nines using hands.
For 6 x 9: Fold down the sixth finger (1st finger on right hand). This partitions the fingers into 5 fingers and 4 fingers which is, of course, the answer to 6 x 9.
- Pose the problem for the students to work on.
- Focus questions that can help students get started include:
How can we set this up?
What information do we know?
What mathematical knowledge could we apply to this situation? Would algebra help?
How will we compare the cases? - Encourage students to clearly justify their reasoning by writing a concluding statement to explain their answer.
- Share answers.
- If the students have not chosen to solve the problem algebraically you could help them get started by expressing one of the child’s postcards for them, for example, Sally has 4s + 3 cards.
Solution
This may be solved using multiples or with algebra and a table.
Option 1: Use a table
Sally has three more than a multiple of 4; Jane has one less than a multiple of 5; and Paul has a multiple of a natural number. If we draw up a table we can compare the numbers of cards for each of Sally and Jane. When we find two numbers are equal we can then see if Paul could have that amount. Since Paul (and therefore everyone else) has less than 100 cards we only have to go up as far as 100 in the table. (In the table 74 gives multiples of 4 and 75, multiples of 5.)
|
M4 |
4 |
8 |
12 |
16 |
20 |
24 |
28 |
32 |
36 |
40 |
44 |
48 |
|
Sally |
7 |
11 |
15 |
19 |
23 |
27 |
31 |
35 |
39 |
43 |
47 |
51 |
|
M5 |
5 |
10 |
15 |
20 |
25 |
30 |
35 |
40 |
45 |
50 |
55 |
60 |
|
Jane |
4 |
9 |
14 |
19 |
24 |
29 |
34 |
39 |
44 |
49 |
54 |
59 |
|
M4 |
52 |
56 |
60 |
64 |
68 |
72 |
76 |
80 |
84 |
88 |
92 |
96 |
|
Sally |
55 |
59 |
63 |
67 |
71 |
75 |
79 |
83 |
87 |
91 |
95 |
99 |
|
M5 |
65 |
70 |
75 |
80 |
85 |
90 |
95 |
100 |
||||
|
Jane |
64 |
69 |
74 |
79 |
84 |
89 |
94 |
99 |
(Incidentally, you might like to think about using a spreadsheet to construct the table.)
We have highlighted the numbers that could represent the number of cards that the children have. There are five possibilities.
But there is one condition that we have not used yet. Paul has an equal number of post cards on every page. Now we assume that his project book has more than one page. Hence the number of post cards must be a multiple of the number of pages. Since he has more than one post card to a page, the number of post cards that they each have is not a prime number. Hence he cannot have 19, 59, or 79 cards. This leaves two possibilities – 39 and 99.
- Suppose that they have 39 cards each. Then since 39 = 4 x 9 + 3, Sally has 9 pages in her book. Since 39 = 5 x 8 – 1, then Jane has 8 pages in her book. Now 39 = 3 x 13, so Paul either has 3 pages with 13 cards on each page or 13 pages with 3 cards on each page. Unless Paul’s project book is pretty big, it’s unlikely that he could get 13 cards to a page. So he probably has 13 pages in his book.
- Suppose that they have 99 cards each. Then since 99 = 4 x 24 + 3, Sally has 24 pages in her book. Since 99 = 5 x 20 – 1, then Jane has 20 pages in her book. Now 99 = 3 x 33 or 9 x 11, so Paul either has 3 pages with 33 cards on each page or 33 pages with 3 cards on each page or 9 pages with 11 cards on each page or 11 pages with 9 cards on each page. Unless Paul’s project book is pretty big, it’s unlikely that he could get as many as 9 cards to a page. So he probably has 33 pages in his book.
Option 2: Use algebra
Here we let s, j and p represent the number of pages in the project books of Sally, Jane and Paul, respectively. So we know that Sally has 4s + 3 cards, Jane has 5j – 1 and Paul has cp, where c is some whole number bigger than 1.
Since Sally and Jane have the same number of cards, 4s + 3 = 5j – 1.
Rearranging gives 5j = 4(s + 1). Since 4 goes into the right-hand side of the equation it must go into the left-hand side. This forces j to be a multiple of 4. Let j = 4k. So 5j – 1 = 20k – 1. Since the children have less than 100 cards, 20k – 1 ≤ 100 and so k ≤ 5 which means that k = 1, 2, 3, 4 or 5.
We can then summarise the possibilities in the table below.
|
k |
1 |
2 |
3 |
4 |
5 |
|
j |
4 |
8 |
12 |
16 |
20 |
|
5j - 1 |
19 |
39 |
59 |
79 |
99 |
|
4s + 3 |
19 |
39 |
59 |
79 |
99 |
|
s |
4 |
9 |
14 |
19 |
24 |
As before (see Option 1) we know that we can discount 19, 59, and 79. The answer now follows as in Option 1.
| Attachment | Size |
|---|---|
| Displaying Postcards.pdf | 55.86 KB |
| Displaying Postcards Maori.pdf | 66.43 KB |
Similar Resources
Difference Magic Squares
This is a problem from the number and algebra strand.
Weighing Time
Solve simultaneous equations
Devise and use problem solving strategies to explore situations mathematically (be systematic, use equipment, draw a diagram, use smaller cases, use algebra).
The Magic Squares
Use algebra as required
Construct magic squares
More Dartboards
Identify and describe patterns in numbers
Work systematically to find the possible outcomes
Triangular Number Links
Use algebra to simplify expressions
Use geometry to assist their algebra.
