Can 5 ^ 23-5 ^ 21 be divided by 120?

Can 5 ^ 23-5 ^ 21 be divided by 120?

5^23-5^21
=5^21*5^2-5^21
=5^21*(5^2-1)
=5^21*24
=5^20*5*24
=120*5^20
5 ^ 20 is an integer, so it's OK

The characteristics of divisible numbers by 4,6,7,8,9,11,13

4: The last two can be divided by four
6: The sum of the numbers is a multiple of three and even
8: The last three can be divided by eight
11: Example: 121
The difference between odd and even digits (greatly reduced) is a multiple of eleven

The characteristic that an n-bit integer can be divided by 3,4,6,7,8,9 respectively
For example: if a three digit number can be divided by three, then the three digit number can be divided by three

4. If the last two digits of an integer can be divided by 4, then the number can be divided by 4
6. If an integer can be divided by 2 and 3, then the number can be divided by 6
7. If the number of one digit of an integer is truncated, and then two times of the number of one digit is subtracted from the remaining number, if the difference is a multiple of 7, then the original number can be divided by 7. If the difference is too large or it is difficult to see whether it is a multiple of 7 by mental arithmetic, we need to continue the above process of "truncation, multiplication, subtraction and difference checking" until we can make a clear judgment. For example, the process of judging whether 133 is a multiple of 7 is as follows: 13-3 × 2 = 7, So 133 is a multiple of 7; for example, the process of judging whether 6139 is a multiple of 7 is as follows: 613-9 × 2 = 595, 59-5 × 2 = 49, so 6139 is a multiple of 7, and so on
8. If the last three digits of an integer can be divided by 8, then the number can be divided by 8
9. If the number sum of an integer can be divided by 9, then the integer can be divided by 9