### Why do this problem?

This problem has several different solutions. The problem can be solved using an experimental / trial and error approach but some consideration of the structure can lead to more efficient solution techniques. There will be no need for students to feel 'stuck' on this problem: they will always be able to experiment with new combinations.

### Possible approach

This printable worksheet may be useful: Consecutive Seven

Students could all write down the numbers $0$ to $20$. One student could be asked to select a first triple and everyone writes that down. All students search for a second triple, whose sum is one more than the first's sum. One such triple is chosen, and everyone writes it down and starts to search for the next - until the task of finding triples whose sums are consecutive is fully understood by the group, at which point, they can work alone or in pairs to find a solution.

With the whole group, ask students to describe what problems occur and how they are dealing with them. Ask them to share any observations, or inspirations they have had. Check that the points in the key questions have been covered in the students comments.

Students who wish to continue to work experimentally could be encouraged to devise a clear recording system for the combinations they are trying. For example, starting with the 20 and 19, what are the possibilities for the other two cards. Students who want to work analytically may choose to use algebra to determine the smallest consecutive number.

### Key questions

• Why have you run out of possibilities? Can you change anything to avoid that problem?
• What are the biggest and smallest sums we could get from a triple?
• Can you work out what the sevenconsecutive numbers will have to add up to?
• Can you select your triples in a logical/symmetrical way?

### Possible extension

• How many different solutions do you think that there might be? Can you work out how many might be possible?
• If the numbers $0$ to $20$ were changed to different sets of $20$ numbers, would solutions still be possible? (specifically designed sets e.g. $\{10, 11, .., 30\}$ or $\{1000, 1001, ...1020\}$ or $\{0, n, 2n, 3n, 4n, ..., 20n\}$, or totally random sets)

### Possible support

It might be helpful to provide students with cards labelled $0$ to $20$ to allow them to make their arrangements. You could also provide calculators so that students can focus on the structure of the problem, rather than getting stuck with the additions.

Alternative questions include:
Can you find seven pairs of numbers which add up to consecutive numbers?
Can you find seven sets of three cards which add up to the SAME number?
Students could write such questions for each other, working in pairs to ensure that these are phrased as accurately as possible.