The law of numbers divisible by 7

The law of numbers divisible by 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
For integer n = ma = 10 * m + A, a is a bit. If m-2a = 7b, it can be divided by 7
Because:
m-2a=7b
m=7b+2a
N=10*m+a
=10(7b+2a)+a
=70b+21a
=7 * (10 * B + 3a), that is, n is divisible by 7

Verification: 1 + 2 + 2 ^ 2 + 2 ^ 3 + +2 ^ (5n-2) can be divisible by 31 (n ∈ n)

It's not 5n-2, it's 5n-1
1+2+2^2+2^3+… +2^(5n-1)
=2^5n-1
=32^n-1
=(31+1)^n-1
=31^n+C(n,1)31^(n-1)+C(n,2)31^(n-2)+...+C(n,n-1)31+1-1
=31^n+C(n,1)31^(n-1)+C(n,2)31^(n-2)+...+C(n,n-1)31
It's obviously a multiple of 31

Verification: 1 + 2 + 2 ^ 2 + +2 ^ (5N -- 1) can divide 31 n into natural numbers

I think this problem can be proved by binary system
31 to binary is 11111
1+2+2^2+…… +2 ^ (5N -- 1) to binary is (11111... 1) 5N ones;
Obviously, the latter can divide the former