35. Find the smallest natural number ending in the digit 6 that quadruples when its last digit is moved to the beginning of the number.
35. First solution. Let the number $X$ satisfy the conditions of the problem:

$$
X=\widehat{a_{1} a_{2} \ldots a_{n-1} 6} \quad \text { and } \quad 4 X=\widehat{6 a_{1} a_{2} \ldots a_{n-1}},
$$

where $a_{1}, a_{2}, \ldots, a_{n-1}, 6$ are the digits of the number $X$. Since the last digit of $X$ is 6, the last digit of $a_{n-1}$ of the number $4 X$ is 4; thus, $X=\overline{a_{1} a_{2} \ldots a_{n-2} 46}$, which allows us to find the second-to-last digit $a_{n-2}=8$ of the number $4 X$. Writing $X$ now as $\ldots 846$, we can find that $a_{n-3}=3$, and so on. We will continue this process until we get the digit 6 in the number $4 X$, which can be considered as carried from the end of the number $X$. Thus, we find that the smallest possible number satisfying the condition of the problem is $X=153846$; in this case, $4 X=615384$.

Second solution. Since the desired number $X$ ends with the digit 6, it can be written in the form $X=10 x+6$, where $x$ is the number obtained from $X$ by erasing the last digit 6. If the number $x$ is $n$-digit, then from the condition of the problem it follows that

$$
4 \cdot(10 x+6)=6 \cdot 10^{n}+x
$$

i.e.,

$$
39 x=6 \cdot\left(10^{n}-4\right), \text { or } \quad x=\frac{2\left(10^{n}-4\right)}{13}
$$

The number $10^{n}-4$ is obviously 6, or 96, or 996, or $9996, \ldots$ The smallest of such numbers that is a multiple of 13 is the number $99996=13.7692$; it corresponds to the value $n=5$. But in this case, the equality (*) gives $x=15384$, and thus, $X=153846$.