How one should approach a problem involving alteration in digits of integers I'm having a lot of difficult to solve problems like these two above:
"Find all integers with initial digit $6$ wich have the following property, that if this initial digit is deleted, the resulting number is reduced to $ 1/25 $ of it's original value."
"Prove that there does not exist an integer which is doubled when the initial digit is transferred the end."
etc.
I know that these problems are very different and may require different approaches but how should one begin to tackle problems like these? any tip will help thx in advance.
 A: As said in the comments, probably the most straightforward approach would be to use the decimal expansion of the numbers in question. This allows you to analyze specific digits. I'm not claiming that this will always work, but it serves as a useful starting point, and it turns out it works here.
As an example for the first one. An integer $k$ with initial digit $6$ is of the form:
$$ 
k = 6\cdot 10^{n - 1} + m
$$
where $n$ is the number of digits in $k$ and $m$ is some integer with $n - 1$ digits. I've written $k$ in this way because I am interested in what happens when the $6$ is removed, and writing it in this way allows me to analyze that because $m$ is precisely the number that I have once the $6$ is removed. I know that when the $6$ is removed the resulting number $m$ is $1/25$ of the original. Thus:
$$
m = \frac{6\cdot 10^{n - 1} + m}{25}
$$
Rearranging we have
$$
4m = 10^{n - 1}
$$
So, $4m$ needs to be a power of $10$, and this power of $10$ should have the same number of zeros as digits in $m$. Now you just need to write out what these numbers are, and then put a $6$ in front of them. The rest of the answer is below.

 For $4m$ to be a power of $10$ we see that $m = 25, 250, 2500, \ldots$. Moreover $4\cdot 25 = 100$. We see that $100$ has two zeros and $25$ has two digits. This works with the others in the list as well so the integers you are looking for are $625, 6250, 62500, \ldots$

