Fear, from the seven digits $0,1,2,3,4,5,6$, many seven-digit numbers without repeated digits can be formed, among which some are multiples of 55. Among these multiples of 55, find the largest and the smallest. (Write out the reasoning process)
Five, Solution: Let the sum of the four digits in the odd positions of such a seven-digit number be $A$, and the sum of the three digits in the even positions be $B$, then $|A-B|=11k$ (where $k$ is a natural number or 0).
Also, $A+B=21$,
Since $A, B$ are positive numbers less than 21, so $A-B=5$, thus $A=16, B=5$.

Among these seven numbers, the sum of three numbers being 5 can only be 0,1,4 and 0,2,3, so,
either (1) $B=0+1+4, A=2+3+5+6$
or (2) $B=0+2+3, A=1+4+5+6$
Since 0 is always in the group of three numbers (i.e., in the even positions),

0 cannot be the last digit, because the number sought is a multiple of 55, so the last digit must be 5. Thus, the smallest number is 1042635. The largest number is 6431205.