Say something enlightening and do the math yourself.

56=7*8

So it can be divisible by 7 and 8 respectively.

A number divisible by 7.

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.

The last three digits of a number divisible by 8 are divisible by 8.

56-5-6=44

The rest of you count what 44 is.

9*5=45

So at least 7 is

Set to x 1 x2x3 x4 x556 (the subscript "1"after x is a number).

So X5 *100+5 *10+6 = 8 * y (a number set by y).

If one digit of an integer is truncated, subtract twice this digit from the rest. If the difference is a multiple of 7, the original number can be divisible by 7. If the difference is too big or it is not easy 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 make a clear judgment.

Use this method to process x 1x2x3x4x556, and see if it can be divisible by 7.

If not, add one more bit and set it to x 1x2x3x4x5x656.

etc

I'll leave the job to you.

By the way, it's hard to call so many.

Add more.

Also, I think I won the first Olympic prize in those days.

O(∩_∩)O~