Find the number of permutations $x_1, x_2, x_3, x_4, x_5$ of numbers $1, 2, 3, 4, 5$ such that the sum of five products \[x_1x_2x_3 + x_2x_3x_4 + x_3x_4x_5 + x_4x_5x_1 + x_5x_1x_2\]
is divisible by $3$.
Since $3$ is one of the numbers, a product with a $3$ in it is automatically divisible by $3,$ so WLOG $x_3=3,$ we will multiply by $5$ afterward since any of $x_1, x_2, \ldots, x_5$ would be $3,$ after some cancelation we see that now all we need to find is the number of ways that $x_5x_1(x_4+x_2)$ is divisible by $3,$ since $x_5x_1$ is never divisible by $3,$ now we just need to find the number of ways $x_4+x_2$ is divisible by $3.$ Note that $x_2$ and $x_4$ can be $(1, 2), (2, 1), (1, 5), (5, 1), (2, 4), (4, 2), (4, 5),$ or $(5, 4).$ We have $2$ ways to designate $x_1$ and $x_5$ for a total of $8 \cdot 2 = 16.$ So the desired answer is $16 \cdot 5=\boxed{080}.$
~math31415926535
~MathFun1000 (Rephrasing for clarity)