**Problem.** We have real numbers such that if we remove one number (any one), then either the sum of the remaining ones is zero or you can split the remaining ones into two sets having equal sum. Prove that all these numbers are zero.

**Solution.** Let be the numbers. Observe that the conditions in the problem can be written as the following system of equations

where Let be the matrix of the coefficients of the above system and let be the column with the entries We can re-write the above system as The problem is asking us to show that So we need to prove that is invertible, i.e. To do this, we use the determinant formula

where is the group of permutations of the set and So is the set of derangements of But we know that the number of derangements of a set with even number of elements is odd. So is a sum of odd number of terms where each term is Clearly this sum can never be zero and hence is invertible.