41. a) Find a six-digit number that increases by 6 times when the last three digits of the number, without changing their order, are moved to the beginning of the number.

b) Prove that there does not exist an eight-digit number that increases by 6 times when the last four digits are moved to the first four positions while maintaining their order.
41. a) Let the number formed by the first three digits of the desired number $N$ be denoted by $p$, and the number formed by the last three digits of $N$ be denoted by $q$. In this case, the condition of the problem gives

$$
1000 q + p = 6(1000 p + q) \quad(=6 N)
$$

from which we get

$$
(1000 q + p) - (1000 p + q) = 999(q - p) = 5 N
$$

i.e., $N$ is divisible by 999.

Further, $p + q = (1000 p + q) - 999 p = N - 999 p$, from which it follows that $p + q$ is also divisible by 999. But $p$ and $q$ are three-digit numbers, which obviously cannot both equal 999; therefore,

$$
p + q = 999
$$

Now we easily find:

$$
(1000 q + p) + (1000 p + q) = 1001(p + q) = 7 N
$$

and, thus,

$$
7 N = 999999, \quad N = 142857
$$

b) Analogously to the solution of part a), denoting by $p$ the number formed by the first four digits of the desired number $N$, and by $q$ the number formed by the last four digits, we find:

$$
7 N = 10001(p + q) = 99999999
$$

which is impossible for an integer $N$ (since 99999999 is not divisible by 7).