What is the maximum possible value of the sine of a $2^{n}$ degree angle, if $n$ is a positive integer?
In the solution, we omit the degree symbol. $\sin \left(2^{n}\right)=\sin \left(2^{n}-360 k\right)$, if $k$ is an integer, and for a suitable $k$, $0 \leq$ $2^{n}-360 k<360$.

In the interval $[0,360)$, within $[0,180]$, the value of $\sin x$ is non-negative, and here it is larger the closer $x$ is to 90. Thus, we seek the values of $n$ and $k$ for which $0 \leq 2^{n}-360 k<360$ and $2^{n}-360 k$ is as close as possible to 90; that is, $0 \leq 2^{n-3}-45 k<45$, and the deviation of $2^{n-3}-45 k$ from $90 / 8=11.25$ is minimal.

The remainders of $2,2^{2}, 2^{3}, \ldots, 2^{12}, 2^{13}$ when divided by 45 are, respectively: $2,4,8,16,32,19,38,31,17,34,23,1,2$. From here on, the remainders repeat periodically, so the remainder closest to 11.25 is 8. Thus, when $2^{n-3}=2^{3}$, $n=6$, and the maximum value of $\sin \left(2^{n}\right)$ is $\sin 64^{\circ} \approx 0.8988$.

Kevei Péter (Szeged, Radnóti M. Gymnasium, 12th grade)