∴4d=a or 10+a or 20+a or 30+a (∵4×9=36, so the maximum value of 4d is 36).

And because 4a=d, 1

∴a can only be 1 or 2, and the corresponding D can only be 4 or 8.

When a= 1, 4d=a cannot be established (because there is no 1 multiple of 4).

So a can only be 2, then 4d=a holds, only d=8.

Then 4c+3=b or 10+b or 20+b or 30+b (3 is 4d decimal place).

Because 4a=d

Therefore, 4b+X=c (x is the decimal place of 4c+3, at this time, c must be less than or equal to 9, otherwise it will not be established if it is rounded to thousands).

∫0≤c≤9，0≤b≤9，0≤X≤3

∴b can only be 0, 1, 2.

Therefore, it is necessary to satisfy that the bit of 4c+3 is equal to b, and the bit of 4b+X is equal to c.

B can only be 1 bit (because 1-3=2, and 0-3=3 and 2-3= 1 are not in the multiple range of 4).

So c can only be 2 or 7 (because 8+3= 1 1).

If c=2

2 128×4=85 12, which is invalid.

So c can only be 7.

2 178×4=87 12

So, this four-digit number is 2 178.