\documentclass[12pt]{article}
\title{Balancing Numbers Problem}
\author{Andrew Poppick}
\usepackage{amsfonts}
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\maketitle

Suppose we have thirteen complex numbers with the following property: if we remove any one of the numbers, the remaining twelve can be split into two sets of 6 numbers each with equal sum.\\

Claim: all thirteen numbers are equal.\\

Proof:
%INTEGERS
\section{Integers}

I will first prove the claim for integers.

Assume $x_1$,...,$x_{13}$ $\epsilon$$\mathbb{Z}$ such that not all $x_i$ are equal.

Let $S=2r_i+x_i$, where $r_i$ is the sum associated with removing an $x_i$

Clearly, S is either even or odd.\\

S is odd $\Rightarrow$$x_i$ is odd, since $2r_i$ is even.

S is even $\Rightarrow$$x_i$ is even, for the same reason.\\

So, either all the $x_i's$ are even, or they are all odd.

Now define $y_i=x_i-x_1$.
Observe that the above property still holds for $y_i$.  Furthermore, since $y_1=x_1-x_1=0$ and $0$ is even, $y_i$ is even for all i.

Now define $y_i'=y_i/2$.  The same property still holds and since $0/2=0$, the numbers are still even.  It is now clear that this process can be repeated indefinitely.  
However, if any nonzero integer is divided through by two enough times, we will come up with an odd number.

$\Rightarrow\Leftarrow$.  Since one of the numbers is always zero, all of the numbers should always be even.  So all of the numbers must be zero.

$y_i=x_i-x_1=0 \Rightarrow x_i=x_1$ for all i.  ie. all of the numbers are equal.
%RATIONALS
\section{Rationals}
We have now proved the claim for integers.  The case for rationals will follow directly from this result.
Assume that $p_1/q_1,...,p_{13}/q_{13}$ $\epsilon\mathbb{Q}$ such that they have the property in question.  Then clearly $z_i=(p_i/q_i)(q_1...q_{13})$ is an integer.  We can apply the above argument to the $z_i's$ to find that $z_1=z_2=...=z_{13}$  But $z_i/(q_1...q_{13})=p_i/q_i$.  So all the $p_i/q_i$'s are equal.
% REALS
\section{Reals}
Now we must prove the claim for the reals.  Assume $r_1,...,r_{13}\epsilon\mathbb{R}$ such that they have the property in question.  $\exists$$q_i\epsilon\mathbb{Q}$ such that $q_i\epsilon(r_i-\epsilon, r_i+\epsilon), \forall$$\epsilon, r_i$\\ 

Furthermore, by Dirichlet's Theorem on simultaneous diophantine approximation, $\forall$ $1\leq i\leq 13$ and $\epsilon>0, \exists$ integers $p_i$ and $q<1/(\epsilon^{13})$ such that $|r_i-p_i/q|<\epsilon/q$ 

ie. all of the reals can be approximated by rationals of a common denominator.\\

If $-6\epsilon/q<(r_1+...+r_6)-\frac{p_1+...+p_6}{q}<6\epsilon/q$

and $-6\epsilon/q<(r_7+...+r_{12})-\frac{p_1+...+p_{12}}{q}<6\epsilon/q$

and $r_1+...+r_6=r_7+...+r_{12}$, then:\\

$-12\epsilon/q<\frac{(p_7+...p_{12})-(p_1+...+p_6)}{q}<12\epsilon/q$

$\Rightarrow -12\epsilon<(p_7+...p_{12})-(p_1+...+p_6)<12\epsilon$\\

But the $p_i$'s are integers.  So set $\epsilon=1/12$  Then it is clear that $(p_7+...+p_{12})-(p_1+...+p_6)=0$\\

ie. $(p_7+...p_{12})=(p_1+...+p_6)$ and we  would come up with something equivalent if we had "picked out" any $r_i$.

$\Rightarrow p_1=p_2=...=p_{13}$\\

So $\forall\epsilon>0$, $|r_i-p/q|<\epsilon/q$ and $p/q$ is the same for each $r_i$\\

$\Rightarrow r_1=r_2=...=r_{13}$

%COMPLEX
\section{Complex Numbers}
The case for the complex numbers follows directly from the reals.  Assume that $c_k=x_k+iy_k\epsilon\mathbb{C}$ for $1\leq$$k\leq$$13$ such that $c_k$'s have the property in question.  

If $c_1+...+c_6=c_7+...+c_{12}$, then:\\

$x_1+...+x_6=x_7+...+x_{12}$, and

$y_1+...+y_6=y_7+...+y_{12}$\\

$\Rightarrow$ all of the $y_i$'s are equal and all of the $x_i$'s are equal.

$\Rightarrow$ all of the $c_i$'s are equal.  QED\\

Note furthermore that this proof did not rely on the number thirteen other than its being an odd number.  The claim should therefore be true for any group of n numbers where n is odd that have the property in question.


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