First Round of the 73rd Czech and Slovak Mathematical Olympiad (take-home part)

a) Yes, the number $N=137658942$ works, since the sums of consecutive triples are (left to right) 11, 16, 18, 19, 22, 21, 15.

b) We shall prove that no such number exists by showing that for any $N$, the sum of the 7 numbers is at most 122, while the numbers given to us have sum 123. Denote the $i$-th digit of $N$ by $a_i$ and let $S$ be the sum of the resulting 7 sums. Then we have: $$\begin{array}{rl}S& =(a_1+a_2+a_3)+(a_2+a_3+a_4)+(a_3+a_4+a_5)+(a_4+a_5+a_6)+(a_5+a_6+a_7)+(a_6+a_7+a_8)+(a_7+a_8+a_9)\\ & =3(a_2+a_3+a_4+a_5+a_6+a_7+a_8)+a_1+a_9\end{array}$$ Since all digits are distinct and non-zero, $a_1,\dots ,a_9$ is a permutation of $1,2,\dots ,9$, so their sum is $45$. Thus, $$S=3(a_2+a_3+a_4+a_5+a_6+a_7+a_8)+a_1+a_9=3(45-a_1-a_9)+a_1+a_9=135-2a_1-2a_9$$ The minimum possible value of $a_1+a_9$ is $1+2=3$, so the maximum possible $S$ is $135-2\times 3=129$. The maximum $a_1+a_9$ is $8+9=17$, so the minimum $S$ is $135-2\times 17=101$.

But more precisely, since the sequence $11,15,16,18,19,21,23$ sums to $123$, which is not possible as $S$ can be at most $122$ (as shown in the original solution), there is no such $N$.