Permutations of seven numbers such that none of the sums are divisible by 3 How many permutations $(a_1,\dots, a_7)$ of $(2,3,4,5,6,7,8)$ exist such that none of the sums
$a_1$,
$a_1+a_2$,
...
$a_1+a_2+\dots a_7$
are divisible by $3$?
I tried to study the cases of the possible $3$-remainders for members of such permutations, but I found that there are just too many possibilities. I hope there is some simple idea that I just can't see.

We can't start with a number that is congruent $0 \mod 3$.  so we must start with one that is congruent $\pm 1 \mod 3$.  If we start $1 \mod 3$ we can't have a $-1 \mod 0$ before we have the second $1\mod 3$.  If we start with a $1 \mod 3$ then have some zeros modulo three, another $1 \mod 3$ we will have no more $1\mod 3$s, so we will have to have, at some point, two $-1 \mod 3$s which will add to $0$.
So the first number must be congruent $-1 \mod 3$, then we must have another $-1 \mod 3$ before we have a $1 \mod 3$.  After the two $-1 \mod 3$s we can't have a third $-1\mod 3$ until we have a $1 \mod 3$.  Once we have two $-1 \mod 3$s and one $1\mod 3$ we must have the third $-1 \mod 3$ before the second $1\mod 3$.
So the order must be $-1\mod 3, *, -1\mod 3, *, 1 \mod 3, *, -1\mod 3, *, 1 \mod 3, * $, where $*$ are zero or more $0\mod 3$.  So of the six remaining places, the two $0\mod 3$ can go anywhere.
So there are ${6 \choose 2}$ ways to arrange $(-1\mod 3,0\mod 3,1\mod 3,-1\mod 3,0\mod 3,1\mod 3,-1\mod 3,0\mod 3)$.
$2,5,8 \equiv -1 \mod 3$ and there are $3!$ ways to order those.  $2!$ ways to order $4,7 \equiv 1 \mod 3$ and $2!$ ways to order $3,6\equiv 0 \mod 3$.
So there are ${6 \choose 2}3!*2!*2! = \frac {6!}{4!2!}*6*2*2 = \frac{5*6}26*2*2 =5*6*6*2 =360$ ways to do this.
