The number of ways in which we can arrange the digits 1, 2, 3, …, 9 such that the product of five digits at any of the five consecutive positions is divisible by 7 is

Let an arrangement of 9 digit number be  $x_1{x}_2{x}_3{x}_4{x}_5{x}_6{x}_7{x}_8{x}_9$.

Note that we require product of each of $\left(x_1,x_2,x_3,x_4,x_5\right);\left(x_2,x_3,x_4,x_5,x_6\right);\dots ;\left(x_5,x_6,x_7,x_8,x_9\right)$ is divisible by 7.

This is possible if the 5th digit is 7.

Therefore, we can arrange the 9 digits in desired number of  ways in $8!$ ways.