A 6-digit number starts with 1. If we move this digit 1 from the first position to the last on the right, we get a new 6-digit number that is three times the original number. What is this number?
The problem is to determine the digits $a, b, c, d$ and $e$ such that the number $a b c d e 1$ is three times $1 a b c d e$:

$$
\frac{\times 3}{a b c d e 1}
$$

Initially, we see that $e=7$, and from there we can start discovering each of the digits:

| $1 a b c d 7$ | $1 a b c 57$ | $1 a b 857$ | $1 a 2857$ |
| :---: | :---: | :---: | :---: |
| $\times 3$ | $\times 3$ | $\times 3$ | $\times 3$ |
| $a b c d 71$ | $a b c 571$ | $a b 8571$ | $a 28571$ |

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