A six-digit number starts with 1. If we move this digit 1 from the first to the last position on the right, we get a new six-digit number, which is three times the original number. What is this number?
The problem is to determine the digits $b, c, d, e$ and $f$ such that the number $b c d$ and $f 1$ is three times $1 b c d e f$.

At the beginning, we see that $f=7$, and from there, we can discover each of the digits, as follows.

![](https://cdn.mathpix.com/cropped/2024_05_01_666a1aec7e447b0a6cbeg-087.jpg?height=166&width=1256&top_left_y=1776&top_left_x=434)

Therefore, $b=4$ and the starting number is 142857.