3. Determine a six-digit number whose products with the numbers $2,3,4,5,6$ in some order are six-digit numbers obtained from the given number by cyclic permutation of the digits.
Solution. Let $N$ denote the desired number. Since $6N$ is also a six-digit number, the first digit of $N$ is 1. The digit 1 is the last digit of one of the numbers $2N, 3N, 4N, 5N$, and $6N$. Clearly, the numbers $2N, 4N, 5N$, and $6N$ cannot have the digit 1 in the units place, so it must be the number $3N$. The number $3N$ has the digit 1 in the units place if and only if the digit in the units place of the number $N$ is 7. Therefore:

$$
\begin{aligned}
& N=1 \_---7 \\
& 3N=\_--\_71
\end{aligned}
$$

If we continue this multiplication: multiplying the second-to-last digit of the number $N$ by 3 should give us a number that ends in 5 (to get the second-to-last digit 7 of the number $3N$ when adding the carry of 2) - this is only possible if that digit is 5, so the digit 5 is the third digit of the number $3N$. Continuing this process, we get $N=142857$ and $3N=428571$. We are left to verify that in this case, the numbers $2N, 4N, 5N, 6N$ are indeed obtained from the number $N$ by cyclic permutation of the digits, which we leave to the reader.