Help with congruences and positive divisors I am studying for an algebra final exam and would like to check if the way I solved this exercise is ok. 
I have been asked to: 

Find all the positive divisors of $25^{70}$ with remainder $2 \pmod 9$ and with remainder $3 \pmod{11}$.


I started by noticing that I can write $25^{70}$ as $5^{140}$ and that finding all the positive divisors means to consider all the $n$ such that $n|5^{140}$. 
I can write those as $n=5^\alpha$ with $0\leq\alpha\leq140$
This means that there are $141$ positive divisors.
Then, I wrote:
$5^{\alpha} \equiv2\pmod{9}$
$5^{\alpha} \equiv3\pmod{11}$.
After that, I checked all the $\alpha$ that satisfy $5^{\alpha} \equiv2\pmod{9}$ and those that satisfy $5^{\alpha} \equiv3\pmod{11}$ simultaneously.
I noticed that for the first term of the system, those such $\alpha$ are $\alpha \equiv5\pmod{6}$, and for the second term they are $\alpha \equiv2\pmod{5}$
To finish, I considered a new system:
$\alpha \equiv5\pmod{6}$
$\alpha \equiv2\pmod{5}$.
And concluded that, by Chinese Remainder Theorem, those are the $\alpha \equiv17\pmod{30}$. Since we said that $0\leq\alpha\leq140$, those such $\alpha$ must be $\alpha=\{ {17, 47, 77, 107, 137}\}$
There are two things I am not sure about: the first one is the way I discovered those $\alpha$. I have to consider too many of them (they are $141$), but I ended up realizing 'by hand' that the cycle was shorter. Am I doing something wrong? Is there another method?
The other thing I am not quite sure about is if I am using properly the conditions I have been given with the positive divisors.
I would appreciate any help.
Thanks.
 A: Your solution is correct. Below, I've provided some further explanation as to how exactly it works. 
Say we want to find all $\alpha\in\mathbb{N}$ with $5^\alpha\equiv 2\pmod 9$. Given any two $\alpha,\beta\in\mathbb{N}$ with $5^\alpha\equiv 2\pmod 9$ and $5^{\alpha+\beta}\equiv2\pmod 9$,we have $5^\beta\equiv 1\pmod 9$. On the other hand, if given some $\alpha,\beta\in\Bbb{N}$ with $5^\alpha\equiv 2\pmod 9$ and $5^{\beta}\equiv 1\pmod 9$, we have $5^{\alpha+\beta}\equiv 2\pmod 9$.
Also, if we have two $\alpha,\beta$ with $5^\alpha\equiv 1\pmod 9$ and $5^{\beta}\equiv 1\pmod 9$, we would have $5^{\gcd(\alpha,\beta)}\equiv 1\pmod 9$. Therefore there exists some $d\in\mathbb{N}$ such that $5^\alpha\equiv 1\pmod 9$ if and only if $d\mid \alpha$.
Combining these results gives that $5^\alpha\equiv 2\pmod 9$ if and only if $\alpha \equiv \beta\pmod d$ for some $\beta$ with $5^\beta\equiv 2\pmod 9$.
We find that when working modulo $9$, we can take $d=6$ (simply the smallest positive integer $d$ such that $5^d\equiv 1\pmod 9$) and an example of a solution would be $5$. Putting $d=6$ and $\beta=5$, this gives:
$$\alpha\equiv 5\pmod 6$$
You can do the same for the other congruence. 
A: Usin the Chinese Remainder theorem, we have to solve for 
\begin{cases}5^n\equiv  2\mod9,\\5^n\equiv  3\mod11.\end{cases}
You won't have to examine so many cases, since by Euler's theorem, $5$ has order a divisor of $\varphi(9)=6$ modulo $9$, and a divisor of $\varphi(11)=10$ modulo $11$. 
Calculating the successive powers of $x$ mod. $9$ and mod. $11$, we see that $5$ has indeed order $6$ mod $9$, but order $5$ mod $11$, so you only have to calculate $30$ modular powers of $5$, and as you found the values $2$ and $5$, you obtain
$$\begin{cases}
5^n\equiv2\mod 9\phantom{1}\iff n\equiv 5\mod6,\\5^n\equiv 3\mod 11\iff n\equiv 2\mod 5.\end{cases}$$
The solutions result from a Bézout's relation between $6$ and $5$, especially simple here: $\;6-5=1$, so the solutions satisfy 
$$n\equiv 2\cdot 6-5\cdot 5=-13\mod \operatorname{lcm}(6,5)=30.$$
The smallest positive value is $17$, and the solutions at most equal to $140$ are
$$\{17,47,77,107,137\}.$$
