# A set of numbers such that when you remove one number, the rest can be arranged into two equal groups with the same sum

March 28, 2012 Leave a comment

Suppose you have a set of 17* *numbers with the property that if you remove any one of the numbers, the remaining can be split into two groups of 8 which sum to the same total. We will show that the numbers must all be equal.

If we have a set of 17 numbers with this property, then we can create another set of 17 numbers by subtracting a constant. So if are one group of numbers with the property, then so is .

Similarly, if have the property, then so do , where *K* is a constant.

In particular, this means that we can shift it so that one of the numbers is zero.

To begin, let us assume the numbers in question are all integers. Let us take out the number which is zero. Therefore we get two groups which both sum to the same (integral) value, call it *S*. Therefore the sum of all the numbers *T* is simply *2S + 0*. Therefore *T* is even.

Let us now take out another one of the numbers, call it *k*. So we get two groups of numbers again which sum to the same value, call it *S’*. So *T = 2S’ + k*. This implies that *k* is even, as *T *is.

Therefore each of the numbers is even. But if that is the case, we can divide all the numbers by two to get another set of integers with the property. This process can be continued indefinitely, which is impossible unless all the numbers were zero to start with. Thus we conclude all the numbers were equal to zero. That means in the original case, before a constant was subtracted, they must have all been equal. (c.f. Proof by Infinite descent)

The case for rational numbers is simple, as you may just multiply all the rational numbers be a constant in order to make them all integers (for instance, by multiplying by the product of the denominators). Thus they are all equal.

The case for real numbers requires a trick. Denote the 17 numbers by . These numbers generate a finite dimensional vector space over the rational numbers (of dimension at most 17, call it *N*). Therefore we can choose a basis (without the Axiom of Choice). Call this basis .

Therefore each of the can be written uniquely as a sum of the multiplied by a rational number. Write for the coefficient of in this representation.

Fix *j*. Then the set of 17 numbers form a set of rational numbers with the property that taking out any one of them allows the remaining to be rearranged into two groups with the same sum. This is because the set of 17 real numbers have this property, and whatever the two sums are, the coefficients must be equal by the uniqueness of basis representations.

However, this implies by the above that the value of is constant for different values of *i*.

Therefore all the coordinates are individually constant, and so the 17 numbers are all equal.

This problem is from *Problems and Theorems in Classical Set Theory*. Obviously, no use is actually made of the fact there are 17 numbers. In fact, no use is made of the fact that the two groups of numbers are equal sizes.