Show that if any ten of eleven integers can be split into two sets with the same total, then the eleven integers must be equal. If one is given eleven integers, and told that for any ten of them it is possible to split them into two sets of five with the same total value, show that the eleven integers must be equal.
My thoughts so far: 
-if you had 10 equal numbers and one different one, you could swap the different one in so it wouldn't work
-if you had nine equal numbers, then clearly those nine plus a different number couldn't be divided up into two equal sets
-if there were eight equal numbers, then calling the equal numbers N and the other numbers $n_1$, $n_2$, $n_3$. If we exclude $n_3$, then either 


*

*$N<n_1,n_2$ or $N>n_1, n_2$, in which case we would require $4N + n_1 = 4 N + n_2$ which implies that $n_1=n_2$. By swapping $n_3$ for $n_1$, we can see that this also implies $n_1=n_2=n_3$. However, then the case of 7$N$s and 3$n$s wouldn't be able to be split into two equal sets. 

*$n_1<N<n_2$, in which case $5N = 3N + n_1 + n_2$ so $2N=n_1 +n_2$. Similarly $N=n_2+n_3=n_1+n+n_3$. However, this isn't possible because either $n_3>N$ in which case $n_2 + n_3 > N$ or $n_3<N$ in which case $n_1+n_3<N$. 
 A: We can prove this by looking at the remainders of the numbers when divided by powers of two.
I first claim that all the numbers must be either even or odd. We start by setting up a split, so that $n_{11}$ is not used, and  $\lbrace n_1, n_2, n_3, n_4, n_5\rbrace$ and $\lbrace n_6, n_7, n_8, n_9, n_{10}\rbrace$ have the same sum. Now suppose there's a mix of odd and even numbers. WLOG let $n_{11}$ and $n_{1}$ have different parities, and swap out $n_1$ so that it is not used any more. We then have two classes $\{n_2,n_3,n_4,n_5,n_{11}\}$ with parity $P_1$ and $\lbrace n_6, n_7, n_8, n_9, n_{10}\rbrace$ with parity $P_2$, and we seek to create a split of these ten numbers so that the parities of the two groups are the same. But that's impossible: any new split can be made from a series of pairwise element swaps, but swapping elements of the same parity leaves both parities unchanged, while swapping elements of opposite parities flips both parities. Therefore, we cannot have a mix of even and odd numbers, and all the numbers must have the same parity.
I further claim that if all the numbers are in the same residue class modulo $p$, they must also be in the same residue class modulo $2p$. Proof: Working modulo $2p$, all the numbers must belong to one of two classes, namely $k$ or $k+p$, for $k \in \{0,\ldots,p-1\}$. As before, we can start with an initially balanced configuration, and swap the spare element with one from the opposite class. Let $S_1$ be the sum of the first group of elements (modulo $2p$), and let $S_2$ be the sum of the second group. Then $S_1 - S_2$ is invariant under pairwise swaps of elements. Swapping elements $n_i$ from group one and $n_j$ from group two makes the sums $$S_1' = S_1 + n_j - n_i\quad \textrm{and} \quad S_2' = S_2 - n_j + n_i$$ and the difference 
\begin{align}
S_1' - S_2' &\equiv S_1 - S_2 + 2(n_j - n_i) \mod 2p\\
&\equiv S_1 - S_2 + 2(k + ap - k + bp) \qquad a,b\in\{0,1\}\\
&\equiv S_1 - S_2 + 2(a-b)p  \\
&\equiv S_1 - S_2
\end{align}
No matter what swaps we make, we can never balance the two sums if we start with unbalanced sums. We must therefore have all elements in the same residue class modulo $2p$.
Combining these facts means that we have
$$ n_i \equiv n_j \mod 2^m$$
for all $i, j \in \{1,\ldots,11\}$ and for all $m\in \mathbb N$. Picking $m$ such that $2^m > \max(n_i)$ lets us conclude that $n_i = n_j$.
A: Suppose there were a counterexample with integers $x_0,x_1,x_2,\ldots,x_{10}$. Without loss of generality, we can assume that $\gcd(x_0,x_1,x_2,\ldots,x_{10})=1$, which implies the $x_i$'s are not all even. But we can also assume that the smallest $x_i$ is $0$ (since $\{x_0-k,x_1-k,x_2-k,\ldots,x_{10}-k\}$ would be another counterexample for any integer $k$). Let's say $x_0=0$ and $x_{10}$ is odd.  But then either $\{x_1,x_2,\ldots,x_{10}\}$ or $\{x_0,x_1,x_2\ldots,x_9\}$ has an odd number of odd integers, which makes it impossible to split into two groups with equal sum. (However you split a set with an odd number of odd integers into two subsets, one subset will have an even sum and the other subset will have an odd sum.)  Thus there can be no counterexample.
Remark: The observation that we can assume the smallest $x_i$ is $0$ is where the "two sets of five" condition is required.
Credit where credit is due: This answer owes a lot to Sten's observation that "all the numbers must be either even or odd." As soon as I read that I saw that the $\gcd=1$ and $x_0=0$ assumptions would produce a contradiction.
