0 roses $\rightarrow$ £56 left - even so can be spent on carnations

1 rose $\rightarrow$ £53 left - odd so cannot be spent on carnations

2 roses $\rightarrow$ £50 left - even so can be spent on carnations

3 roses $\rightarrow$ odd amount left - not good

4 roses $\rightarrow$ even amount left - good

etc

Only even numbers of roses until the maximum possible number of roses

£56 $\div$ 3 = 18 remainder 2 (1 carnation)

Maximum 18 roses

10 even numbers (including 0)

10 possible bunches

0 carnations $\rightarrow$ £56 left but 56 is not a multiple of 3

1 carnation $\rightarrow$ £54 left $\rightarrow$ 54$\div$3 = 18 roses

2 carnations $\rightarrow$ £52 left but 52 is not a multiple of 3

3 carnations $\rightarrow$ £50 left but 50 is not a multiple of 3

4 carnations $\rightarrow$ £48 left $\rightarrow$ 48$\div$3 = 16 roses

etc

3 more carnations $\Rightarrow$ £6 less to spend on roses = 2 fewer roses

Every 3rd carnation gives a possible number of roses

Maximum number of carnations: 56$\div$2 = 28

1, 4, 7, 10, ..., 28 allowed

2 less than the 3 times table, up to 30

so there are 10 possible bunches (since 30 = 3$\times$10)

You can find more short problems, arranged by curriculum topic, in our short problems collection.