Question 57
A permutation of the integers $1901, 1902, \ldots, 2000$ is a sequence $a_{1}, a_{2}, \ldots, a_{100}$ in which each of those integers appears exactly once. Given such a permutation, we form the sequence of partial sums
$$
s_{1} = a_{1}, \quad s_{2} = a_{1} + a_{2}, s_{3} = a_{1} + a_{2} + a_{3}, \ldots, s_{100} = a_{1} + a_{2} + \cdots + a_{100}.
$$
How many of these permutations will have no terms of the sequence $s_{1}, \ldots, s_{100}$ divisible by three?
$$
s_{1} = a_{1}, \quad s_{2} = a_{1} + a_{2}, s_{3} = a_{1} + a_{2} + a_{3}, \ldots, s_{100} = a_{1} + a_{2} + \cdots + a_{100}.
$$
How many of these permutations will have no terms of the sequence $s_{1}, \ldots, s_{100}$ divisible by three?