Smallest natural number $n$ that fulfills given conditions Find the smallest natural number $n$ such that the following conditions are fulfilled:


*

*It's decimal representation ends with the number 6.

*If the 6 is moved to the beginning of the number, the resulting number is 4 times the original.


I am not sure where to begin with such a question. Any ideas?
 A: Well, if your $n$ ends with 6, then you may write it as $10m+6$. Also, suppose it was $k$ digits long; then with 6 moved to the beginning it would be $6\cdot10^k+m$. The rest is simple:
$$4(10m+6)=6\cdot10^k+m\\
39m=6\cdot(10^k-4)\\
m=2\cdot(10^k-4)/13\\$$
...and we only have to check a handful of values for $k$ to find the smallest one that fits.
A: Another way to arrive at the answer is to write it as a multiplication problem
$$\begin {align} *\ * \ * \ * \ *\ *\ 6& \\ \underline {\quad \quad \quad \quad \quad \quad \quad 4}&\\  6\ * \ * \ * \ *\ *\ * \end {align}$$ where the stars represent digits and we don't know how many there are.  We can see that the leading digit of the multiplicand is a $1$ because otherwise the leading digit of the product is greater than $6$, and the trailing digit of the product is a $4$, so put them in.$$\begin {align} 1\ * \ * \ * \ *\ 4\ 6& \\ \underline {\quad \quad \quad \quad \quad \quad \quad 4}&\\  6\ 1 \ * \ * \ *\ *\ 4 \end {align}$$  Now the tens digit of the product is an $8$.  Keep going until you get a $1$ in the multiplicand and you are there.
$$\begin {align} 1 \ 5 \ 3 \ 8\ 4\ 6& \\ \underline {\quad \quad \quad \quad \quad 4}&\\  6 \ 1 \ 5 \ 3\ 8\ 4 \end {align}$$
