I thought of a six-digit number. I erased its first digit and wrote it at the end, and thus I got three times the original number. Which number did I think of?
Solution. The original number $A=10^{5} a+10^{4} b+10^{3} c+10^{2} d+10 e+f$, and the number $B$ obtained by moving the first digit to the last place is $B=10^{5} b+10^{4} c+10^{3} d+10^{2} e+10 f+a=3 A=3\left(10^{5} a+10^{4} b+10^{3} c+10^{2} d+10 e+f\right)$; from this,

$$
299999 a=7 \cdot 10^{4} b+7 \cdot 10^{3} c+7 \cdot 10^{2} d+7 \cdot 10 e+7 f
$$

that is,

$$
42857 a=10^{4} b+10^{3} c+10^{2} d+10 e+f
$$

a number with at most five digits. Thus, the value of $a$ (non-zero) can be 1 or 2 (with 3, we would get a six-digit number).

If $a=1$, then the number in question is 142 857, and if $a=2$, then $42857 \cdot 2=85714$, so the number in question is 285714. It is easy to verify that these numbers meet the requirements.