<s>[INST] Mandelbrot Competition Problem Please give only hints for this problem:
When $(x^2+x+1)^{54}$ is expanded, how many coefficients of the resulting polynomial are divisible by 3?
So far I tried rewriting $(x^2+x+1)^{54}= (\frac{x^3-1}{x-1})^{54}$ to try to get two binomials and use binomial expansion, but since this is a quotient, I am stuck. Please give hints and the highest level of math required to take the Mandelbrot Competition, which this problem is from, is precalculus.
 [/INST] Note that 
\begin{align}
(1+x+x^2)^3 \mod 3 &= (x^{6} + 3 \; x^{5} + 6 \; x^{4} + 7 \; x^{3} + 6 \; x^{2} + 3 \; x + 1) \mod 3\\
&= 1+x^3 +x^6
\end{align}
Using the above with $x$ replaced by $x^3$, we get 
$$(1+x+x^2)^9 \mod 3 = (1+x^3+x^6)^3 \mod 3 = 1+x^9 + x^{18}$$
and $$(1+x+x^2)^{27} \mod 3 = (1+x^9+x^{18})^3 \mod 3 = 1+x^{27}+x^{54}$$
Thus
\begin{align}
(1+x+x^2)^{54} \mod 3 &= (1+x^{27}+x^{54})(1+x^{27}+x^{54}) \mod 3 \\
&= x^{108} + 2 \; x^{81} + 3 \; x^{54} + 2 \; x^{27} + 1\\
&= x^{108} + 2 \; x^{81}  + 2 \; x^{27} + 1
\end{align}
Thus there are only 4 terms that are not multiples of 3. The rest of 105 terms are multiples of 3.
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