Problem
Mary had a $5 note. She exchanged it with her brother Sam for some coins.
Could she have exchanged it for exactly 6 coins?
What is the largest number of coins that Sam could have given Mary?

What is this problem about?
This problem not only shows knowledge of money and simple arithmetic, but it also gives children the opportunity to make systematic lists and explore number.

Achievement Objectives
Measurement (Level 2)
- represent a sum of money by two or more different combinations of notes and coins

Mathematical Processes
- devise and use problem solving strategies to explore situations mathematically (systematic list, use equipment)

Resources
play money
Blackline master of the problem (English)
Blackline master of the problem (Maaori)

Specific learning outcomes
The children will be able to:
- represent a sum of money by a combinations of coins
- use a list to work systematically

1. Play guess the note ($5). Give a number of clues and ask the children to guess the note that you are describing, for example.
   This note has a ...on it.
   This note is not blue
2. Read the problem to the class.
3. Brainstorm for ways to solve the problem.
4. As the children work on the problem ask questions that encourage them to work systematically:
   Tell me how you are solving the problem? Why did you select that strategy?
   Are there other coins that you could have used?
   How do you know that you have checked all the possibilities?
5. Share solutions.

Extension to the problem
Could Mary have exchanged the $5 for 10 coins?

Solution
There are at least two answers to question A. The coins might have been $2, $2, 50c, 20c, 20c and 10c, or $1, $1, $1, $1, 50c and 50c. But are they all the possible answers?   Don't expect the children to give such a complete justification as the one that follows.  The answer with a description of how they got it is fine at this level.

To do this systematically takes a little time. Clearly there can be at most two $2 coins. So suppose that there are precisely two $2 coins. Then we have to make up $1 with four coins. We can’t have two 50c coins but we can have one. Then we have to have at least one 10c coin. Two 20c coins give the only way to finish this case off to give $2, $2, 50c, 20c, 20c, 10c.

What if we only have one $2 coin? Then we can have two, one or no $1 coins. We can’t have two $1 coins because we would have to be able to make up $1 with three coins and this isn’t possible. One $1 coin would mean that we would have to find four coins to make up $2. This can only be done with four 50c coins. So another answer is $2, $1, 50c, 50c, 50c, 50c. So suppose that we have no $1 coins. We have to make up $3 with five coins. But the biggest coin we can have now is 50c and five 50c just isn’t big enough. This means that there is only one answer with one $2 coin.

The most $1 coins we can use is four. This leads to $1, $1, $1, $1, 50c, 50c. If we only use three $1 coins then we have to make up $3 with three coins but 3 x 50 isn’t big enough. The same goes for fewer than three $1 coins. So there is only one way to use $1 coins (and no $2 coins). It looks as if there are  three answers to problem A.

Part B. The most coins Sam could have would be when he only had 5c coins. There are a hundred fives in 500. So he would have a hundred 5c coins.

Extension. This is clearly a lot harder than the 6 coin problem. Maybe one way to handle this is to keep a list of the answers that the children give and see how many they can come up with.

Since 10 x 50 = 500, Sam had to have had at least one coin worth 50c or more. So we have to look systematically at the cases where Sam has $2 coins, $1 coins and 50c coins. The only answers are

$2, $2, 50c, 20c, 5c, 5c, 5c, 5c, 5c, 5c;

$2, $2, 50c, 10c, 10c, 10c, 5c, 5c, 5c, 5c;

$2, $2, 20c, 20c, 10c, 10c, 10c, 10c, 10c, 10c;

$2, $1, $1, 50c, 20c, 10c, 5c, 5c, 5c, 5c;

$2, $1, $1, 50c, 10c, 10c, 10c, 10c, 5c, 5c;

$2, $1, $1, 20c, 20c, 20c, 20c, 20c, 5c, 5c;

$2, $1, $1, 20c, 20c, 20c, 10c, 10c, 10c, 10c;

$2, $1, 50c, 50c, 50c, 10c, 10c, 10c, 10c, 10c;

$2, $1, 50c, 50c, 20c, 20c, 20c, 20c, 10c, 10c;

$2, 50c, 50c, 50c, 50c, 20c, 20c, 20c, 20c, 20c;

$1, $1, $1, $1, 50c, 20c, 10c, 10c, 5c, 5c;

$1, $1, $1, $1, 50c, 10c, 10c, 10c, 10c, 10c;

$1, $1, $1, $1, 20c, 20c, 20c, 20c, 10c, 10c;

$1, $1, $1, 50c, 50c, 50c, 20c, 10c, 10c, 10c;

$1, $1, $1, 50c, 50c, 20c, 20c, 20c, 20c, 20c;

$1, $1, 50c, 50c, 50c, 50c, 50c, 20c, 20c, 10c;