11.1. In a six-digit number, the first digit, which is 2, was moved to the last place, leaving the other digits in the same order. The resulting number turned out to be three times larger than the original. Find the original number.
Answer: 285714.

Solution. According to the condition, the desired number has the form $\overline{2 a b c d e}$, then we have: $\overline{a b c d e 2}=$ $3 \cdot \overline{2 a b c d e}$, or $\overline{a b c d e} \cdot 10+2=3 \cdot(200000+\overline{a b c d e})$. Let $\overline{a b c d e}=X$ - a five-digit number, then $10 X+2=3 \cdot(200000+X)$, or $7 X=600000-2, X=85714$, then the original number will be $\overline{2 a b c d e}=\overline{2 X}=285714$.