Can you draw any conclusions?
We found out that you can always do odd numbers, multiples of 3,5 and 7. I also think that all numbers except for powers of two are possible (as long as they have odd factors).
Can you support your conclusions with convincing arguments or proofs? (formulae and explanations).
You can always do odd numbers because if you choose an odd number (n) add one (n-1) then divide by 2 ((n+1)÷2) so then you take the answer (a) and add 1 so (a+1) then the answer and the answer plus one equals your number (a+(a+1)=n) so the formula would be (n-1)÷2=a then (a+1)+a=n. You can do multiples of 3, 5&7 all in a similar way you just change the number of consecutive numbers depending on what the number is. First choose a number (n) then divide that number by 3 (or 5 or 7) (n÷3 or replace the 3 with a 5 or 7) then that number is the number in the middle so if you have 3 you would have your answer -1 then your answer then your answer +1 ((a- 1)+a+(a+1)=n) but if you have 5 you would have to -2 -1 a +1 +2 and for 7 -3 -2 -1 a +1 +2 +3 so the formula would be n÷3=a then (a-1)+a+(a+1)=n. And so on for 5 and 7. This might also work for 9, 11 and all the odd multiples but not the even multiples. I think it is only the powers of two that are not possible because they have no odd factors so long as the multiple has odd factors it is possible. I have not proven this though.
What are the possible next questions?
Perhaps something like what is the smallest number with 19 consecutive numbers going into it.
Possible next questions relating to this could be what is the smallest number with 19 consecutive numbers going in to it. Or you could experiment with dividing and multipling or subtracting the numbers. Also you could try consecutive multiples, such as 3+6+9 or 4+8+12