## First Solution.

Assume that such a two-digit number exists; that is, assume that the number $10+x$ (where $x$ is a single-digit number or 0) satisfies the conditions of the problem. Then $10 x+1=3 \cdot(10+x)$, which leads to the equation $7 x=29$ which has no solutions in the set of natural numbers.

If we assume that such a three-digit number $100+x$ exists, we get the condition $10 x+1=3 \cdot(100+x)$ and the equation $7 x=299$ which also has no solutions in the set of natural numbers.

For a four-digit number, we get the condition $10 x+1=3 \cdot(1000+x)$, and the equation $7 x=2999$ which also has no solutions in the set of natural numbers.

For a five-digit number, we get the condition $10 x+1=3 \cdot(10000+x)$, and the equation $7 x=29999$ which also has no solutions in the set of natural numbers.

For a six-digit number, from the condition $10 x+1=3 \cdot(100000+x)$, we get the equation $7 x=299999$, whose solution is $x=42857$.

The number 142857 is the smallest number that satisfies the conditions of the problem.

## Note:

"The problem can also be solved by determining one by one the digits in the product

$1 \ldots 7 \cdot 3=\ldots 71$. 1
## 1. task

Find the smallest multi-digit number that starts with the digit 1 and has the property that, if we move this digit 1 from the beginning to the end, behind the units digit, we get a number three times larger than the original.