Question 133
$n$ is a positive integer and relatively prime with $6$. $a_1, \dots, a_n, b_1, \dots, b_n$ are positive integers such that $a_1 \lt a_2 \lt \dots \lt a_n$ and $b_1 \lt b_2 \lt \dots \lt b_n$. If for arbitrary natural number $t$ such that $a_i + a_j + a_k = t$ ($i \lt j \lt k$) triple's number is equal to $b_i + b_j + b_k = t$ ($i \lt j \lt k$) then prove that $a_1 = b_1, a_2 = b_2, \dots, a_n = b_n$.
(proposed by G. Batzaya)
(proposed by G. Batzaya)