Probability of a sum being divisible by $4$ Roll $n$ six-sided dice.
What is the probability that the sum of the results is divisible by $4$?
This is an 11th grade problem, and I really don't know -after a lot of searching-what tools do I need to solve this problem using only high school knowledge.

I'm not quite sure that generating functions and some complex numbers are included in 11th grade (not sure what it corresponds to in Sweden), but you can get a closed solution for each value of $n$ if you know about these. Even though I suspect this is too complicated.
Let the generating function be
\begin{equation}
f_n(x) = (x+x^2+x^3+x^4+x^5+x^6)^n
\end{equation}
For two dices, expanding $f_2(x)$ gives
\begin{equation}
f_2(x) = x^2+2x^3+3x^4+4x^5+5x^6+6x^7+5x^8+4x^9 +3x^10+2x^{11}+x^{12}
\end{equation}
here, it is the coefficients of $x^4,x^8$ and $x^{12}$ which are interesting, the sum of these corresponds to the number of outcomes which are divisible by four. Summing up, we see that for two dices, there are $3+5+1=9$ ways to get a sum divisible by four.
Now, define the number $\zeta = e^{\pi i/2}$. This has a periodicity of $4$, e.g. $\zeta^1 = \zeta^5$. If you expand $f_n(x)$, you get $f_n(x) = \sum_{k=0}^{6n} c_kx^k$, so here we want to try to extract the coefficients of interest, i.e. $c_4,c_8,c_{12}, ...$.
If you now look at the function $g_k(x)=x^k, k=0,1,2,3,...$, note that
\begin{align} 
g_0(\zeta^0) + g_0(\zeta^1) + g_0(\zeta^2) + g_0(\zeta^3) &= 4 \newline
g_1(\zeta^0) + g_1(\zeta^1) + g_1(\zeta^2) + g_1(\zeta^3) &= 0 \newline
g_2(\zeta^0) + g_2(\zeta^1) + g_2(\zeta^2) + g_2(\zeta^3) &= 0 \newline
g_3(\zeta^0) + g_3(\zeta^1) + g_3(\zeta^2) + g_3(\zeta^3) &= 0 \newline
g_4(\zeta^0) + g_4(\zeta^1) + g_4(\zeta^2) + g_4(\zeta^3) &= 4 \newline
\end{align}
and so on. As you can see, we extract the correct coefficients. This means that we have that the number of possible ways $N$ to have a sum divisible by four is given by
\begin{equation}
N = \frac{1}{4} \sum_{k=0}^3 f_n(\zeta^k) 
\end{equation}
Calculating gives
\begin{align}
f_n(\zeta^0) &= 6^n \newline
f_n(\zeta^1) &= (-1+i)^n \newline
f_n(\zeta^2) &= 0 \newline
f_n(\zeta^3) &= (-1-i)^n
\end{align}
such that
\begin{equation}
N = \frac{1}{4} \sum_{k=0}^3 f_n(\zeta^k) = \frac{1}{4} \left(6^n + 2^{n/2+1} \cos\left(\frac{3\pi n}{4} \right) \right)
\end{equation}
The first few terms give that the number of total outcomes which are divisible by four are then $N=1,9,55,322,1946$ for $n=1,2,3,4,5$. These are the same I got when expanding the generating functions so I think things match up, even though I'm seldom certain.
Furthermore, since you asked about the probability, the total number of outcomes are $6^n$, the probability for the sum being divisible by four, is given by
\begin{equation}
P_n = \frac{1}{4} \left(1 + \frac{2^{n/2+1}}{6^n} \cos\left(\frac{3\pi n}{4} \right) \right)
\end{equation}
Finally, note that $P_n\rightarrow 1/4$ when $n\rightarrow \infty$.
