Problem:
Let f(x)=log10(sin(πx)⋅sin(2πx)⋅sin(3πx)⋯sin(8πx)). The intersection of the domain of f(x) with the interval [0,1] is a union of n disjoint open intervals. What is n?
Answer Choices:
A. 2
B. 12
C. 18
D. 22
E. 36
Solution:
Let g(x)=sin(πx)⋅sin(2πx)⋅sin(3πx)⋯sin(8πx). The domain of f(x) is the union of all intervals on which g(x)>0. Note that sin(nπ(1x))=(−1)k+1sin(nπx),\
so g(1x)=g(x). Because g(1/2)=0, it suffices to consider the subintervals of (0,1/2) on which g(x)>0. In this interval the distinct solutions of the equation g(x)=0 are the numbers k/n, where 2≤n≤8,1≤k<n/2, and k and n are relatively prime. For n=2,3,4,5,6,7, and 8 there are, respectively, 0,1,1, 2,1,3, and 2 values of k. Thus there are 1+1+2+1+3+2=10 solutions of g(x)=0 in the interval (0,1/2). The sign of g(x) changes at k/n unless an even number of factors of g(x) are zero at k/n, that is unless there are an even number of ways to represent k/n as a rational number with a positive denominator not exceeding 8. Thus the sign of g(n) changes except at 1/4=2/8 and 1/3=2/6.
Let the solutions of g(x)=0 in the interval (0,1/2) be x1,x2,…,x10 in increasing order, and let x0=0 and x11=1/2. It is easily verified that x5=1/4 and x7=1/3, so for 0≤j≤10, the sign of g(x) changes at xj except for j=5 and 7 . Because 5 and 7 have the same parity and g(x)>0 in (x0,x1), the solution of g(x)>0 in ( 0,1/2) consists of 6 disjoint open intervals. The solution of g(x)>0 in (1/2,1) also consists of 6 disjoint open intervals, so the requested number of intervals is 12.
The problems on this page are the property of the MAA's American Mathematics Competitions