Explanation :

Let us consider (for clarity), the number to be PQRSTUVW.

So, ST must be divisible by 5.
--> As no zero is allowed only possibility is T = 5.
--> U must be 2 or 6 as TU is divisible by 4.

Now, PQ must be divisible by 8, and QR by 7.
--> PQ can be 16, 24, 32, 48, 64, 72.
--> QR can be 14, 21, 28, 42, 63.
--> Clearly, PQR can be 163 or 321 or 328 or 721 or 728.
--> Corresponding possible values of PQRS for above PQR values are : 1636, 3218, 3284, 7218, 7284.
--> Clearly 1636 is impossible, 'coz of repetition of 6. So, we are left with the following values of PQRSTU yet : 321856, 328456, 721856, 728456.

Now, UV is divisible by 3. As, U can only be 6 (as seen from above list), only 63 is the possible case.
So, we can reduce the possible set to : 7218563, 7284563 now.
The only thing we are to do now is to make the number even, by adding a digit at the right. We can add only 2, 4, 8 or 6.
In the first case, we can use only 4, 'coz all others are used up.
In the second, we cannot use any of them, so it can be rejected.

We are finally, left with 72185634.