\section*{Problem 22}

What is the smallest \(\mathrm{n}\) for which there is a solution to \(\sin \mathrm{x}_{1}+\sin \mathrm{x}_{2}+\ldots+\sin \mathrm{x}_{\mathrm{n}}=0, \sin \mathrm{x}_{1}+\) \(2 \sin \mathrm{x}_{2}+\ldots+\mathrm{n} \sin \mathrm{x}_{\mathrm{n}}=100 ?\)


\section*{Solution}

Put \(\mathrm{x}_{1}=\mathrm{x}_{2}=\ldots=\mathrm{x}_{10}=3 \pi / 2, \mathrm{x}_{11}=\mathrm{x}_{12}=\ldots=\mathrm{x}_{20}=\pi / 2\). Then \(\sin \mathrm{x}_{1}+\sin \mathrm{x}_{2}+\ldots+\sin \mathrm{x}_{20}=(-1\) \(-1-1-\ldots-1)+(1+1+\ldots+1)=0\), and \(\sin x_{1}+2 \sin x_{2}+\ldots+20 \sin x_{20}=-(1+2+\ldots+\) 10) \(+(11+12+\ldots+20)=100\). So there is a solution with \(\mathrm{n}=20\). If there is a solution with \(\mathrm{n}\) \(<20\), then there must be a solution for \(\mathrm{n}=19\) (put any extra \(\mathrm{x}_{\mathrm{i}}=0\) ). But then \(100=\left(\sin \mathrm{x}_{1}+2\right.\) \(\left.\sin \mathrm{x}_{2}+\ldots+19 \sin \mathrm{x}_{19}\right)-10\left(\sin \mathrm{x}_{1}+\sin \mathrm{x}_{2}+\ldots+\sin \mathrm{x}_{19}\right)=-9 \sin \mathrm{x}_{1}-8 \sin \mathrm{x}_{2}-7 \sin \mathrm{x}_{3}-\ldots\) \(-\sin \mathrm{x}_{9}+\sin \mathrm{x}_{11}+2 \sin \mathrm{x}_{12}+\ldots+9 \sin \mathrm{x}_{19}\). But \(\mid\) rhs \(\mid \leq(9+8+\ldots+1)+(1+2+\ldots+9)=\) 90. Contradiction. So there is no solution for \(\mathrm{n}<20\).

