Will Rearrangement Apply here?

Actually there's no such problem posted by anybody. It's made by me as a follow-up problem for you.

@Nashita Rahman , @Md Zuhair

There are a good amount of such numbers.

Some examples are:

$20853 \\ 28350 \\ 25830 \\ 23058 \\ 20832 \\ 20538 \\ 32508 \\ 35280 \\ 38052 \\ 58023 \\ 53802 \\ 80325 \\ 82530 \\ 85302$

My approach here was to find the remainders left when a certain place value of the given numbers are divided by $7$ and then use modular arithmetic to devise like this:

$\begin{aligned} \pmod{7} \equiv & \{\phi , 1 , 5 , 6 , 4 \} \to 10^4 \text{ for } \{ 0 , 2,3,5,8 \} \text{ respectively } \\ & \{0,5,4,2,6\} \to 10^3 \text{ for } \{ 0 , 2,3,5,8 \} \text{ respectively } \\ & \{0,4,6,3,2\} \to 10^2 \text{ for } \{ 0 , 2,3,5,8 \} \text{ respectively } \\ & \{0,6,2,1,3\} \to 10^1 \text{ for } \{ 0 , 2,3,5,8 \} \text{ respectively } \\ & \{0,2,3,5,1\} \to 10^0 \text{ for } \{ 0 , 2,3,5,8 \} \text{ respectively } \end{aligned}$

The $\phi$ above denotes that $0$ cannot occupy the $10^4$ place.

Notice that to form numbers divisible by $7$ we have to pick out respective numbers from each set such that no two numbers correspond to the same position in the sets (as no repetition is allowed) and such that the sum of the numbers turns out to be a multiple of $7$ . Since the maximum sum that we can make here is $6+6+6+6+0 = 24$ (without any correspondence), therefore, we have to pick up the numbers that will give $7,14$ or $21$ .

The above mentioned are some of the numbers which I was able to find by some observation. If there's any further shortcut we can apply to find these numbers, please let me know.