1981 AMC 12 Problem 18

Use odd symmetry and first count the positive solutions

Only positive sine humps below $x=100$ can meet the line, with special care for the first hump

The equation is odd-symmetric and has the solution $x=0.$ On $(0,\pi)$ there is one further positive solution. For each $k=1,2,\ldots,15,$ the positive sine hump on $(2k\pi,(2k+1)\pi)$ rises above $\frac{x}{100}$ and then falls below it, giving two solutions. There are no positive solutions beyond $31\pi$, because the next positive hump begins above $100$ while $|\sin x|\le1.$ Thus there are $1+2(15)=31$ positive solutions, the same number negative, and zero itself: $31+31+1=63.$

Therefore, the correct answer is C.