Is there a 6-digit number that becomes 6 times larger when its last three digits are moved to the beginning of the number, keeping the order of the digits?
Let's take a six-digit number, and denote the number formed by its first three digits as $A$, and the number formed by the other three digits as $B$.

According to the conditions of the problem, the following equality must hold:

$$
6(1000 A + B) = 1000 B + A
$$

Rearranging, we get:

$$
857 A = 142 B
$$

Since $A$ and $B$ are integers, and 857 is a prime number, this can only be true if $B$ is divisible by 857. The only three-digit number that satisfies this is $B = 857$, and then $A = 142$.

Thus, there is only one six-digit number that meets the requirements, and that is 142857. Indeed, $142857 \cdot 6 = 857142$.