how to find out remainder of $3^{256}$ divided by $13$ here is a question that about finding the remainder when dividing $3^{256}$ divided by $13$.  Can anyone suggest how to find the solution

In general when the modulus $m$ is small, you can check the successive powers $x^r$ modulo $m$, the remainders will cycle quickly.
For instance let's examine $4^r\pmod 7$
$\begin{cases}
4^0\equiv 1\pmod 7\\
4^1\equiv 4\pmod 7\\
4^2\equiv 16\equiv 2\pmod 7\\
4^3\equiv 4^2\times 4\equiv 2\times 4\equiv 8\equiv 1\pmod 7\\
4^4\equiv 4^3\times 4\equiv 1\times 4\equiv 4\pmod 7\\
\cdots
\end{cases}$
You can see the remainders will always be $1,2,4,1,2,4,1,\cdots$
Thus the result depends of the divisibility of $r$ by $3$. 


*

*If $r=3k$ then $4^r$ will be the same as $4^0\equiv 1\pmod 7$

*If $r=3k+1$ then $4^r$ will be the same as $4^1\equiv 4\pmod 7$

*If $r=3k+2$ then $4^r$ will be the same as $4^2\equiv 2\pmod 7$
For instance $4^{857}\equiv 4^{(857\bmod 3)}\equiv 4^2\equiv 2\pmod 7$

Of course with another base, the cycle may be shorter or longer (e.g. $3^r\pmod 7$ leads to remainders $1,3,2,6,5,4,1,\cdots$
And this time $3^r\equiv 3^{(r\bmod 6)}\pmod 7$
This technique works very well for small modulus, however when $m$ becomes larger the cycle length may grow as well. In some cases, depending on the base, the cycle can still be short, but in general the method become less applicable.
So for bigger modulus, you need to uses more advanced stuff, like little Fermat theorem or Euler's totient function, both allow to reduce the exponent $r$ when $m$ is a prime number (or a power of a prime).
When $m$ is a composed number (of $n$ primes), Chinese remainder theorem, allows to transform the problem into $n$ smaller problems where only primes modulus are involved.

Applying to your problem $3^r\equiv 1,3,9,1,\cdots \pmod{13}$ the cycle is only $3$ steps short.
Thus $3^{256}\equiv 3^{(256\bmod 3)}\equiv 3^1\equiv 3\pmod{13}$
