1) Choose any 6 numbers from first 7 natural numbers, express the other number using these 6 numbers exactly once and only one operator. This should work for any choice of 6 numbers. What is the operator that has to be chosen?
2) Replace 6 by (n-1) and 7 by n in (1), find for which n, this logic works. If the chosen numbers have to be used at least once, find the other n values for which this logic works.
I assume that the one operator can be used multiple times, with parenthesis if needed. It seems obvious that addition and multiplication would not work, and division would work only for some cases, so the simplest operator is then subtraction:
1 = 2-3-4-(6-5-7)
2 = 1-3-4-(5-6-7)
3 = 1-5-(6-2-4-7)
4 = 1-2-3-(5-6-7)
5 = 1-2-3-(4-6-7)
6 = 2-1-(7-3-4-5)
7 = 2-1-(6-3-4-5)
I conjecture that the values of n for part 2 are n mod 4 = 0 or 3 with exception of n=3.
Part 1 Solution
