How many solutions does the equation
sin(x) sin(2x) sin(3x) · · · sin(11x) sin(12x) = 0
have in the interval (0,π] ? Choices:
A) 11
B) 12
C) 24
D) 46
E) 68
If x is a solution of the equation, then lx = k for some integers k and l with 1 _&lt;l_&lt;12. Thus   x = k   lπ. Since we want x to be in (0;π], the number of solutions equals the number of fractions in   (0,1] with numerator an integer, and denominator a positive integer _&lt;12. Since we want to avoid   counting the same solution twice (for example 2π/=3 = 4π/6), we need to count only fractions in   lowest terms. Let φ(n) be the number of fractions in lowest terms in (0,1] with denominator n.   Then φ(1) = 1, φ(2) = 1,φ(3) = 2, φ(4) = 2,φ(5) = 4, φ(6) = 2,φ(7) = 6, φ(8) = 4,   φ(9) = 6, φ(10) = 4,φ(11) = 10, and φ'(12) = 4. Adding the above values of φ we get that the   number of solutions is 46.   correct answer 46
<result>D</result>
How many solutions does the equation
sin(x) sin(2x) sin(3x) · · · sin(11x) sin(12x) = 0
have in the interval (0,π] ?