The number of real solutions to the equation \[ \frac{x}{100} = \sin x \] is

$\text{(A)} \ 61 \qquad \text{(B)} \ 62 \qquad \text{(C)} \ 63 \qquad \text{(D)} \ 64 \qquad \text{(E)} \ 65$
1. Consider the graphs of \( y = \frac{x}{100} \) and \( y = \sin x \) separately. 

2. The function \( y = \sin x \) is periodic with period \( 2\pi \). This means that the sine function repeats its values every \( 2\pi \) units along the x-axis.

3. The function \( y = \frac{x}{100} \) is a straight line with a slope of \( \frac{1}{100} \). This line intersects the sine curve at various points.

4. To find the number of intersections, we need to determine how many times the line \( y = \frac{x}{100} \) intersects the sine curve \( y = \sin x \) within one period of \( 2\pi \).

5. Within one period \( 2\pi \), the sine function oscillates between -1 and 1. The line \( y = \frac{x}{100} \) will intersect the sine curve twice within each period: once when the sine curve is increasing from -1 to 1 and once when it is decreasing from 1 to -1.

6. To find the total number of intersections, we need to determine how many periods of \( 2\pi \) fit within the range where the line \( y = \frac{x}{100} \) intersects the sine curve.

7. The line \( y = \frac{x}{100} \) intersects the sine curve from \( x = -100 \) to \( x = 100 \). This range is \( 200 \) units long.

8. The number of periods of \( 2\pi \) within this range is given by:
   \[
   \frac{200}{2\pi} = \frac{100}{\pi}
   \]

9. Since each period contains 2 intersections, the total number of intersections is:
   \[
   2 \times \frac{100}{\pi}
   \]

10. Using the approximation \(\pi \approx 3.14\), we get:
    \[
    \frac{100}{\pi} \approx \frac{100}{3.14} \approx 31.85
    \]

11. Since we are looking for the number of intersections, we round \( 31.85 \) to the nearest whole number, which is 32.

12. Therefore, the total number of intersections is:
    \[
    2 \times 32 = 64
    \]

13. However, we must consider the intersection at \( x = 0 \) which is counted twice (once for positive and once for negative x-values). Therefore, we subtract 1 from the total count:
    \[
    64 - 1 = 63
    \]

Conclusion:
\[
\boxed{63}
\]