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 $x_1{x}_2{x}_3+x_2{x}_3{x}_4+x_3{x}_4{x}_5+x_4{x}_5{x}_1+x_5{x}_1{x}_2$ is divisible by 3

Let  $x_3=3$ then
$x_1{x}_2{x}_3+x_2{x}_3{x}_4+x_3{x}_4{x}_5+x_4{x}_5{x}_1+x_5{x}_1{x}_2=3\left(x_1{x}_2+x_2{x}_4+x_4{x}_5\right)+x_5{x}_1\left(x_2+x_4\right)$
So, the sum is divisible by 3 if $x_2+x_4$ is dvisible by 3, the possible sums of $\left(x_2+x_4\right)$ can
be (1, 2), (2, 1), (1, 5), (5, 1), $x_1\text{ and }{x}_5$ (2, 4). (4, 2), (4, 5) or (5, 4) are we have 2 ways
to designate for a total of 8 x 2 = 16
So, desired = 16 x 5 = 80