Find all numbers divisible by 25, that begin with 6. please, help me solve this problem:

Find all numbers divisible by 25, that begin with 6.

Regards.
 A: An integer is divisible by $25$ if and only if the last two digits are $00$, $25$, $50$, or $75$. Therefore, there are four possibilities.
$$\pm6X00,\ \pm6X25,\ \pm6X50,\ \pm6X75$$
where $X$ is any sequence of digits, perhaps empty.
(Don't forget the negative integers as well! Though you could quibble about whether a number like $-625$ starts with a minus sign or with a $6$.)
@GeoffRobinson makes the excellent point that this works for decimal, base 10, representation. This would need to be changed for other number systems.
A: They are all numbers of the form
$$6X\dots XYY$$
where $X\dots X$ could be anything, and $YY=00, 25, 50, 75$.
A: I'm going to assume base 10 (decimal), and that they have to be positive.
There are infinitely many such numbers. For any $n > 1$, you can just do $6 \times 10^n$ to get one such number, then keep adding 25 and stop short of $7 \times 10^n$ to get all such numbers with $n + 1$ digits.
Do note that $625 = 25^2 = 5^4$. Are there other powers of 5 that begin with 6? That's a slightly more interesting question. (There's at least one other).
A: If you want use the rules of modular-arithmetic you can write:
$$z=a_0+a_1\cdot 10+...+a_n\cdot 10^n$$
With $a_n=6$ 
The number $z$ is divisible for $25$ if $$a_0+10\cdot a_1$$ is a multiple of $25$
Indeed $$a_0\equiv a_0\pmod {25}$$
$$10\cdot a_0\equiv 10\cdot a_0\pmod {25}$$
While the other digits are divisible for $25$
Indeed $$10^2\equiv 0\pmod {25}$$
And etc...
