Sally, Jane and Paul collect postcards. They each buy a project book to display them in.

The books aren’t necessarily all the same size.

Sally suggests they put the same number of postcards on each page of their own book, and that there should be more than one postcard on each page.

Sally says, "If I place my postcards four to a page, I'll have three cards left over".

Jane says, "If I had just one more postcard I could group mine five to a page".

Paul says, "I can fit mine perfectly on each page already. I guess I've got the most so I'll be the first to get to 100".

But when they count their postcards they find they all have the same number.

How many postcards do they each have?

How many pages do their project books have?

The pattern in this problem is generated from the rule given in words that need careful interpretation. Students work with multiples of 4 and 5, and must exclude prime numbers from their solution.

Using a table is one approach, and an algebraic approach using letters is another.

Of interest: The equations in this problem are known as 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__.)

You may also wish to find out more about *Fermat’s Last Theorem* which involved a Diophantine equation, xn + yn = zn for integers x, y, z, where n is a whole number bigger than 2.

### Problem

Sally, Jane and Paul collect postcards. They each buy a project book to display them in. The books aren’t necessarily all the same size.

Sally suggests they put the same number of postcards on each page of their own book, and that there should be more than one postcard on each page.

Sally says, "If I place my postcards four to a page, I'll have three cards left over".

Jane says, "If I had just one more postcard I could group mine five to a page".

Paul says, "I can fit mine perfectly on each page already. I guess I've got the most so I'll be the first to get to 100".

But when they count their postcards they find they all have the same number.

How many postcards do they each have?

How many pages do their project books have?

### Teaching sequence

- Begin by exploring the pattern of the multiples of 9. Have students discuss these.

For example, the sum of the 2 digits in each multiple equal 9:**18**(1 + 8 = 9),**27**(2 + 7 = 9),**36**(3 + 6 = 9) etc., Ask if this pattern continues. - Pose the postcard 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 solution.
- Share solutions and discuss the strategies students use.
- If the students have not chosen to solve the problem algebraically, model an expression for one of the child’s postcards, for example, Sally has 4s + 3 cards, and have students continue themselves.

### 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.