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## 'Consecutive Seven' printed from http://nrich.maths.org/

### 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

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.