Problem: Let $f(x) = \log_{10} \left(\sin(\pi x) \cdot \sin(2 \pi x) \cdot \sin (3 \pi x) \cdots \sin(8 \pi x)\right)$ . The intersection of the domain of $f(x)$ with the interval $[0,1]$ is a union of $n$ disjoint open intervals. What is $n$
$\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 12 \qquad \textbf{(C)}\ 18 \qquad \textbf{(D)}\ 22 \qquad \textbf{(E)}\ 36$

Solution: You should be able to somewhat visualize what the $\sin$ function looks like (if you can't then you should look it up and try to memorize it). To summarize, the graph of $y= \sin{x}$ is positive from the interval $(0, \pi)$ and negative from the interval $(\pi, 2\pi)$ (notice how the intervals use parentheses instead of brackets, as brackets would denote inclusive bounds and that it incorrect).
All of the various $\sin$ functions that we are multiplying are in the form of $k\pi x$ , which means the intervals have now changed to integer values.
Also, the period of the function $y =\sin{k \pi x}$ is equivalent to $\frac{2}{k}$ ; For example the period of $y=  \sin{4\pi x}$ is $\frac{1}{2}$
Let a "wave" be a full period of a $\sin$ curve. Then, a "half-wave" is half of a full period of a $\sin$ curve. Over the interval $[0, 1],$ $y= \sin{\pi x}$ will display $1$ "half-wave", $y= \sin{2\pi x}$ will display $2$ "half-waves", and $y= \sin{k\pi x}$ will display $k$ "half-waves". Also notice that the graph of $y= \sin{k \pi x}$ crosses the $x$ -axis exactly $k-1$ times over the interval $(0, 1)$
The problem asks for the number of disjoint intervals, not the values of the intervals themselves. This means we only want the positive values of
$\prod_{k=1}^8 \sin{k\pi x}$
(the product of all of the various $\sin$ functions)
We don't care about the specific values themselves, we just need to know what the sign (+/-) of the product is. Notice that if an even number of the functions are negative, the product will be positive; conversely, if the number of negative functions is odd, the product will be negative. Therefore we only need to count the number of intervals during which an even number of functions are negative/positive. By definition, the sign of a function changes after it crosses one of its roots (because trig functions don't have multiplicities). Notice that the roots of $y = \sin{k\pi x}$ are in the form
$\frac{1}{k}, \frac{2}{k}, \frac{3}{k},...$
(this is in agreement with the fact that $y= \sin{k \pi x}$ crosses the $x$ -axis exactly $k-1$ times over the interval $(0, 1)$
Therefore, across the interval $[0, 1]$ , the sign will change every time the function hits a root of one of the $\sin$ functions. We must be careful, however, as some roots (such as $\frac{1}{2}$ and $\frac{2}{4}$ ) are duplicates.
Notice that the duplicates always come in pairs. For example, $\frac{1}{2}$ is also $\frac{2}{4}, \frac{3}{6},$ and $\frac{4}{8}$ ; there are four equivalent roots. Across the interval $[0, 1],$ there are no duplicate roots that come in triplets, or in fives. The roots are always single, doubled up, or quadrupled up. So every time the function reaches a single root, it switches signs, and every time it reaches a duplicate root, the sign remains the same (this still counts as 2 distinct intervals, though, as the zeroes "separate" them). Listing out the roots (not including $x=0$ and $x=1$ ):
$1$ : n/a
$2: \frac{1}{2}$
$3: \frac{1}{3}, \frac{2}{3}$
$4: \frac{1}{4}, \frac{2}{4}, \frac{3}{4}$
$5: \frac{1}{5}, \frac{2}{5}, \frac{3}{5}, \frac{4}{5}$
$6: \frac{1}{6}, \frac{2}{6}, \frac{3}{6}, \frac{4}{6}, \frac{5}{6}$
$7: \frac{1}{7}, \frac{2}{7}, \frac{3}{7}, \frac{4}{7}, \frac{5}{7}, \frac{6}{7}$
$8: \frac{1}{8}, \frac{2}{8}, \frac{3}{8}, \frac{4}{8}, \frac{5}{8}, \frac{6}{8}, \frac{7}{8}$
Then in order (showing duplicates as well):
$\frac{1}{8}, \frac{1}{7}, \frac{1}{6}, \frac{1}{5}, \frac{1}{4}, \frac{1}{4}, \frac{2}{7}, \frac{1}{3}, \frac{1}{3}, \frac{3}{8}, \frac{2}{5}, \frac{3}{7}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{4}{7}, \frac{3}{5}, \frac{5}{8}, \frac{2}{3}, \frac{2}{3}, \frac{5}{7}, \frac{3}{4}, \frac{3}{4}, \frac{4}{5}, \frac{5}{6}, \frac{6}{7}, \frac{7}{8}$
Then creating a sequence of signs based on the roots:
$+,-,+,-,+,+,-,-,+,-,+,+,-,+,-,-,+,+,-,+,-,+$
We see that there are $12$ positive intervals $\Rightarrow \boxed{12}$

Answer: B 12