13.358. Find a six-digit number starting with the digit 1 such that if this digit is moved to the end, the resulting number is three times the original number.
Solution.

The original six-digit number has the form $1 \cdot 10^{5}+x$. After moving the digit 1 to the last place, we get the number $10 x+1$. According to the condition, $10 x+1=3\left(1 \cdot 10^{5}+x\right)$, from which $x=42857$.

Answer: 142857.