First of all, because special numbers are all numbers consisting of 0 and 7, they must all be multiples of 7 (because 0 and 7 are multiples of 7). Since each special number is a multiple of 7, we need to add several special numbers to get 1, so we might as well divide them all by 7, which is equivalent to each number consisting of 0 and 1 (we might as well call such numbers quasi-special numbers).

Now every number is a quasi-special number composed of 0 and 1 If you add them together, you will get 0. 1428 57 142857 ... As far as the numbers in each digit are concerned, they will add up to 1, 4, 2, 8, 5 and 7 respectively. If there is no carry, obviously, every bit is on. Because the largest is 8, there must be at least 8 quasi-special numbers; If carry occurs, it means that the sum of a certain bit is at least 10, and there must be at least 10 quasi-special numbers, which is obviously more.

So, can eight do it? According to the above analysis, it is obviously possible, such as the following eight figures:

0. 1 1 1 1 1 1 1 1 1 1 1 1 ...( 165438)

0.0 1 1 1 10 1 1 1 1 1 1 ...(0 1 165438)

0.010110111... (0101/kloc-

0.010110111... (0101/kloc-

0.000111001... (0001/cycle);

0.0001010010/... (00010/cycle);

0.0001010010/... (00010/cycle);

0.00010000100 ... (000100 cycle);

Calculate the number of 1 on each bit, and we know that their sum is exactly 0. 1428 57 142857. ...

What about the original standard number? These eight numbers are multiplied by seven respectively.

I wish you happiness.