There is a six-digit number, and when it is multiplied by 2,3,4,5,6, the result is still a six-digit number, all the same number

1:
This problem embodies the infinite knowledge in the numbers.
"7" is one of them.
1/7=0.142857 142857 142857 ...
2/7=0.285714 285714 285714 ...
3/7=0.428571 428571 428571 ...
4/7=0.571428 571428 571428 ...
5/7=0.714285 714285 714285 ...
6/7=0.857142 857142 857142 ...
It is so magical that 6 scores are made up of the same "6" numbers, only in the position of numbers.
And the position is not irregular change.
Careful observation will find that "6" the position of the number is just before and after the movement, the slightest confusion.
The multiple relationship between the 6 fractions determines the multiple relationships between the "6" cycles.
So: 142857:285714:428571:571428:714285:857142 =1:2: 3:4: 5:6
Of course, we can also get the results by calculation.
The process is as follows: Suppose there is a "6" bit number, if the maximum number of bits is moved to the bit, the new number is n times the original number. and 2≤n≤6.
First, the highest number of digits is equal to a, the remaining 5 bits are equal to B, multiples are equal to N, and: 2≤n≤6.
Then, the original number can be expressed as 100000a+b;
The new number after displacement can be expressed as 10b+a
Establishing an equation relationship: 10b+a=nx (100000a+b)
by discussion, N = 2, 4, 5, 6 are not available.
N when and only if equals 3 equation can be changed: 10b+a=3x (100000A+B)
7b=2999999a
b=42857a
After discussion, A can only take 1, b get 42857
The original number is: 142857 will move 142857 respectively 1-bit, 2-bit, 3-bit, 4-bit, 5-bit, the new number obtained is not difficult to find their multiple relationship with the original number.
2:
First, assume that the six-digit number represented as ABCdef

According to test instructions, it is known that a,b,c,c,d,e,f are not equal to 6 numbers, and none of them are 0, the highest bit a=1.

The result of the abcdef*5 is that there is a 5 in the last 0 a,b,c,c,d,e, and F is the cardinality. But F is not 5 (because ABCDE5 times 2,4,6 result is 0).

So f=3 or 7 or 9. Multiply the numbers by 1, 2, 3, 4, 5, 6, respectively:

3, 6, 9, 2, 5, 8 (6 different numbers, not counting 1)

7* 7, 4, 1, 8, 5, 2 (including 1 with 6 different numbers)

9* 9, 8, 7, 6, 5, 4 (6 different numbers, not counting 1)

So f=7. And these 6 numbers are 1,2,4,5,7,8.

Because 1bcde7 times 2 is 2****4, and multiplied by 3 is 4****1, so b=4. That is 14cde7.

Then 14cde7 times 2 is 28cde4, so c=2. That is 142de7.

Then there are only two things: 142587 (discard); because 142587*3=427761

or 142857 (correct). Only this one positive solution

There is a six-digit number, and when it is multiplied by 2,3,4,5,6, the result is still a six-digit number, all the same number