3 consecutive summing to 13

 You permute each of the digits $0, 1,2, \dots, 9$ and write them in a single row.

A) Show that no matter what the permutation, some 3 adjacent elements of the row sum to at least $13$.

B) Can you find a permutation where no 3 adjacent elements sum to more than $13$?

Scroll down for solution

.

.

A) Say the permutation is $x_0, x_1, \dots, x_9$.

Now one of $x_0$ or $x_9$ is $< 9$. We can assume $x_0 < 9$.

Let $S_i = x_i + x_{i+1} + x_{i+2}$.

Since $x_0 < 9$ we must have that $S_1 + S_4 + S_7 > 36$ and thus $\text{max} \{S_1, S_4, S_7\} > 12$.

B)  9 3 1 7 4 2 6 0 5 8