Interesting solution...
Number Theory
Level
3
Suppose there are 2N+1 numbers. Any 2N numbers of them can be devided into two groups, each group has N elements, so that the sum of the elements of each group are equal. Can we deduce that all the 2N+1 numbers are equal?
No.
Yes.
It depends on N.
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1 solution
First, we can learn that all 2N+1 numbers are either all even or all odd. If they are all even, devide all of them by 2, and the proposition still remains true. If they are all odd, then subtract 1 from all of them and then devide them by 2, and the proposition still remains true. This will not stop unless at one moment all the numbers turn into zero. So we can proove that they are equal at the beginning.