42. Find a six-digit number, the products of which when multiplied by 2, 3, 4, 5, and 6 are written with the same digits as the number itself, but in a different order.
42. Let $x$ be the number satisfying this condition. Since $6x$, like $x$, is a six-digit number, the first digit of the decimal representation of the number $x$ is 1. Therefore:
1) the first digits of the decimal representation of the numbers $x, 2x, 3x, 4x, 5x$, and $6x$ are all different, so they form a complete set of digits participating in the decimal representation of the number $x$;
2) all digits in the decimal representation of the number $x$ are different from each other.

Among these digits, there is no 0, so the last digit of the number $x$ is odd (otherwise $5x$ would end in zero) and different from 5 (otherwise $2x$ would end in zero). Therefore, the last digits of the decimal representation of the numbers $x, 2x, 3x, 4x, 5x$, and $6x$ are all different, and thus also form a complete set of digits participating in the decimal representation of the number $x$. Therefore, among them there is a 1. The number $3x$ can end in 1, as $2x, 4x$, and $6x$ end in even digits, $5x$ ends in 5, and the decimal representation of $x$ already has a 1 (the first digit). Therefore, $x$ ends in the digit 7, $2x$ in the digit 4, $3x$ in the digit 1, $4x$ in the digit 8, $5x$ in the digit 5, and $6x$ in the digit 2. Since the first digits of these numbers are the digits of the same set, arranged in ascending order, we can write:

$$
\begin{aligned}
& x \cdot 1=1 * * * * 7 \\
& x \cdot 2=2 * * * * 4 \\
& x \cdot 3=4^{* * * * 1} \\
& x \cdot 4=5 * * * * 8 \\
& x \cdot 5=7 * * * * 5 \\
& x \cdot 6=8^{* * * * 2}
\end{aligned}
$$

(stars stand in place of the unknown digits).

Note now that in the written table, not only in each row do the six different digits $1,2,4,5$, 7, and 8 appear in some order, but also in each column, the same six different digits appear in some order. Indeed, suppose, for example, that in the numbers $x \cdot 2$ and $x \cdot 5$ the same digit $a$ (which can have one of two values not occupied by the first and last digits of the two numbers being considered) stands in the third place. In this case, the difference $x \cdot 5 - x \cdot 2 = x \cdot 3$ will represent a six-digit number, in the decimal representation of which the third place will be the digit 0 or the digit 9 (this follows from the rule of subtracting multi-digit numbers "in a column"). But this is impossible, since we know that the number $x \cdot 3$ is written with the digits $1,2,4,5,7$, and 8.

Now let's add the numbers

$$
\begin{aligned}
& x \cdot 1=1 * * * * 7 \\
& x \cdot 2=2 * * * * 4 \\
& x \cdot 3=4^{* * * * 1} \\
& x \cdot 4=5 * * * * 8 \\
& x \cdot 5=7 * * * * 5 \\
& x \cdot 6=8 * * * * 2
\end{aligned}
$$

"column-wise," considering that the sum of the digits in each column is $1 + 2 + 4 + 5 + 7 + 8 = 27$. We get:

$$
x \cdot 21=2999997
$$

or $x=142857$. This is the desired number; indeed,

$$
\begin{aligned}
x & =142857, \\
2x & =285714 \\
3x & =428571, \\
4x & =571428, \\
5x & =714285 \\
6x & =857142
\end{aligned}
$$