[ Decimal numeral system ]

Find the smallest natural number starting with the digit 4 and decreasing by four times when this digit is moved to the end of the number.

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Answer: 410256.

Let the desired number be $\mathrm{X}=\overline{4 a b \ldots c}=4 \cdot 10^{\mathrm{n}}+A$, where $A=\overline{a b \ldots c}$ ( $A$ is an $n$-digit number). After moving the first digit to the end, we get the number $Y=\frac{}{a b \ldots c 4}=10 A+4$. According to the condition,

$$
4 \cdot 10^{\mathrm{n}}+A=4(10 A+4)
$$

from which $39 A=4 \cdot \underbrace{99 \ldots 9}_{n-1} 6$ or $13 A=4 \cdot \underbrace{33 \ldots 33}_{n-1}$ 2. By performing the division by 13 "in column", we find the smallest value of $A: A=4 \cdot 33332 / 13=10256$. Therefore, the smallest value of $X$ is 410256.

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