Problem: Snow White and the Seven Dwarves are living in their house in the forest.
On each of 16 consecutive days, some of the dwarves worked in the diamond mine
while the remaining dwarves collected berries in the forest.
No dwarf performed both types of work on the same day.
On any two different (not necessarily consecutive) days,
at least three dwarves each performed both types of work.
Further, on the first day, all seven dwarves worked in the diamond mine.
Prove that, on one of these 16 days, all seven dwarves were collecting berries.

\end{enumerate}

Solution: Let $Q_n$ denote the vector space $\left\{ 0,1 \right\}^n$.
For a vector $v$, let $v[i]$ denote the $i$th component.
We may identify each day with a vector $v_k$,
where $v_k[i] = 0$ if dwarf $i$ worked in the diamond mine,
and $v_k[i] = 1$ otherwise.
Let $V = \left\{ v_1, v_2, \dots, v_{16} \right\}$
be the subset of $Q_7$ in the problem,
and assume $v_{16} = 0000000$.

We first prove the following:
\begin{lemma*}
  Exactly two vectors start with each of $000$, $001$, \dots, $111$.
  Similar statements hold for any choice of three indices.
\end{lemma*}
\begin{proof}
  If three vectors start with $000$ (say)
  then we run into problems.
\end{proof}

Now we know that $v_{16}$ is the all-null vector.
Ignoring that vector (and hence considering just the first $15$ vectors),
let $n_i$ be the number of $1$'s in $v_i$, where $i = 1, 2, \dots, 15$.
\begin{claim*}
  We have
  \begin{align*}
  \sum_{i=1}^{15} \binom{n_i}{1} &= \frac{16}{2} \binom{7}{1} = 56 \\
  \sum_{i=1}^{15} \binom{n_i}{2} &= \frac{16}{2^2} \binom{7}{2} = 84 \\
  \sum_{i=1}^{15} \binom{n_i}{3} &= \frac{16}{2^3} \binom{7}{3} = 70.
  \end{align*}
\end{claim*}
\begin{proof}
  This follows using the lemma and double counting.
  For example, the third equation is counting
  the number of quadruples $(i, j_1, j_2, j_3)$
  such that the $i$th string has a $1$
  in the $j_1$th, $j_2$th, $j_3$th position, where $j_1 < j_2 < j_3$.
  The left-hand side counts the number when summed by $i$;
  the right-hand side counts the number according
  to the $\binom73$ choices of $(j_1, j_2, j_3)$.
\end{proof}

We may then rewrite the lemma as
\begin{align*}
  \sum_{i=1}^{15} n_i &= 56 \\
  \sum_{i=1}^{15} n_i^2 &= 2 \cdot 84 + 56 = 224 \\
  \sum_{i=1}^{15} n_i^3 &= 6 \cdot 70 + 3 \cdot 224 - 2 \cdot 56 = 980.
\end{align*}
Now remark that $(n_i-3)(n_i-4)(n_i-7) \le 0$ for each integer $1 \le n_i \le 7$.
We compute
\begin{align*}
  0 &\ge \sum_{i=1}^{15} (n_i-3)(n_i-4)(n_i-7) \\
  &= \sum_{i=1}^{15} n_i^3 - 14n_i^2 + 61n_i - 84 \\
  &= 1 \cdot 980 - 14 \cdot 224 + 61 \cdot 56 - 15 \cdot 84 \\
  &= 980 - 3136 + 3416 - 1260 \\
  &= 0.
\end{align*}
This implies that $n_i \in \left\{ 3,4,7 \right\}$ for each $i$.
If $n_i \neq 7$ for any $i$ then we have
\[ 224 = \sum n_i^2 \equiv 15 \cdot 2 \not\equiv 0 \pmod{7} \]
which is impossible.
This means that $n_i = 7$ for some $i$, and therefore, $1111111 \in V$, as desired.

\begin{remark*}
  Up to permutation there turns out to be a unique set of $16$ vectors.
  Xinyang Chen gives the following compact description of the unique solution:
  \begin{itemize}
    \ii $0000000$ and $1111111$, of course
    \ii $1101000$ and its six \emph{cyclic shifts} $0110100$, $0011010$, etc.
    \ii The complements of the above seven strings (i.e.\ $0010111$, $1001011$, etc.).
  \end{itemize}
  To write out all sixteen explicitly as a TL;DR:
  \begin{align*}
    \Big\{ 0000000, 1111111,
      & 1101000, 0110100, 0011010, 0001101, 1000110, 0100011, 1010001 \\
      & 0010111, 1001011, 1100101, 1110010, 0111001, 1011100, 0101110 \Big\}.
  \end{align*}
\end{remark*}