3. Find a six-digit number that, when multiplied by 2, 3, 4, 5, and 6, gives other six-digit numbers composed of the same digits but in a different order.
3. Let the desired number be denoted as: \( N = \overline{a_{6} a_{5} a_{4} a_{3} a_{2} a_{1}} \), with \( a_{6} \neq 0 \), because otherwise \( N \) would not be a six-digit number. \( a_{6} \neq 2 \), because \( 5N \) would then be a seven-digit number. Therefore, \( a_{6} = 1 \).

It is clear that the first digits of the six numbers \( N, 2N, 3N, 4N, 5N \), and \( 6N \) increase, starting from 1. Among them, there is no zero. All these digits are different. \( a_{1} \neq 0 \), \( a_{1} \neq 1 \), because the first digits of all six numbers are different and all digits \( a_{6}, a_{5}, a_{4}, a_{3}, a_{2}, a_{1} \) are distinct. \( a_{1} \neq 5 \). If \( a_{1} = 5 \), then \( 2N \) ends in zero, and zero is not among the desired digits. \( a_{1} \) cannot be even, otherwise \( 5N \) would end in zero. If \( a_{1} = 3 \), then \( N \) ends in 3, \( 2N \) in 6, \( 3N \) in 9, \( 4N \) in 2, \( 5N \) in 5, and \( 6N \) in 8. Considering that \( a_{6} = 1 \), we get 7, not 6 digits. Therefore, \( a_{1} \neq 3 \). If \( a_{1} = 7 \), then the numbers \( N, 2N, 3N, 4N, 5N, 6N \) end in 7, 4, 1, 8, 5, and 2. We get six different digits, and among them is 1. If \( a_{1} = 9 \), then these same numbers would end in 9, 8, 7, 6, 5, and 4. Among them, there is no digit 1, although \( a_{6} = 1 \). Therefore, \( a_{1} \neq 9 \). Thus, \( a_{1} = 7 \).

![](https://cdn.mathpix.com/cropped/2024_05_21_ec9ddf54a7c86775293eg-075.jpg?height=389&width=462&top_left_y=1850&top_left_x=363)

Fig. 10

Summing up, we get \( 21N = 2999997 \), from which \( N = 142857 \). The numbers \( N, 2N, 3N, 4N, 5N \), and \( 6N \) can be easily read using the circle (Fig. 10).

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