You said: twice the odd sum MINUS the even sum. If it is divisible by 7, then this number can be divisible by 7. This is obviously not true. For example, 1005928, the sum of its odd number and two even numbers is 29, which cannot be divisible by 7, but 1005928 can obviously be divisible by 7!

There are two ways to judge whether a number is divisible by 7:

① Tail cutting method:

If one digit of an integer is truncated, twice that digit will be subtracted from the remainder. If the difference is a multiple of 7, the original number can be divisible by 7. If the difference is too large or it is difficult to see whether it is a multiple of 7 in mental arithmetic, you need to continue the above-mentioned process of "rounding, multiplication, subtraction and difference test" until you can clearly judge. 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 another example, the process of judging whether 6 139 is a multiple of 7 is as follows: 6 13-9× 2 = 595, 59-5× 2 = 49, so 6 139 is a multiple of 7, and so on.

Tail cutting method:

Proof process:

Let p = a1+a2 *10+a3 *10 2+...+a (n-1) *10 (n-1)+an *.

q=a2+a3* 10+...+a(n- 1)* 10^(n-2)+an* 10^(n- 1)-2a 1

2p+q=2 1(a2+a3* 10+...+an* 10^(n- 1))

And because 2 1=7*3, if P is a multiple of 7, then we can get that Q is a multiple of 7.

(2) The last three methods:

The difference between the last three digits and the first three digits of this number (or vice versa) can be divisible by 7, 1 1, 13. This number is divisible by 7, 1 1 and 13.

For example: 1005928

Last three digits: 928, first three digits: 1005 1005-928=77.

Because 7 | 77,7 |1005928.

The last three methods are briefly proved:

Let a number be abcdef = ABC×1000+def = ABC×1001-ABC+def = ABC× 7×13×1-(ABC