Saudi Arabian Mathematical Competitions, 2015

We label the vertex of the hexagon by $1,2,3,4,5,6$ and suppose that six numbers are written in the order $a, b, c, d, e, f$ on the edges $(1,2), (2,3), \ldots, (6,1)$, respectively.

We first notice that by choosing a vertex of the hexagon and adding 1 to the two numbers written on two adjacent sides to the vertex, the difference $(a + c + e) - (b + d + f)$ is an invariant. Thus, if we want to obtain a hexagon with equal numbers on its sides, we must have $a + c + e = b + d + f$ at the beginning.

We will show that it is also sufficient. Let $N$ be a big integer, for example, we can take $N > 30$. We choose the vertices $1,2,4,5$ and add 1 to the two numbers written on two sides of those vertices by $N - f, N - b, N - c$ and $N - e$ times, respectively. Then, we obtain a hexagon with six numbers

$a + 2N - b - f, N, N, d + 2N - e - c, N, N$.

Since $a + c + e = b + d + f$, we have $a + 2N - b - f = d + 2N - e - c = M > N$ (since $N > 30 = a + b + c + d + e + f$). We now can choose the vertices 3, 6 for $M - N$ times each, then we obtain a hexagon with numbers $M$ on all of its sides.

Now, we count how many 6-tuples $(a, b, c, d, e, f)$ of distinct positive integers with $a + c + e = b + d + f = 15$.

We list all triples of distinct positive integers with sum 15: $(1,2,12), (1,3,11), (1,4,10), (1,5,9), (1,6,8), (2,3,10), (2,4,9), (2,5,8), (2,6,7), (3,4,8), (3,5,7), (4,5,6)$. We then check that there are exactly 19 times we can pair these triples to obtain 6 distinct numbers.

Since we can permute the numbers on each tuple, the total number of 6-tuples satisfying the given conditions is $19 \times 6!$.