Rich is right: the easiest way to solve this one is to visualize it.
Taking it a step further, the mathematical concept being tested here is DIVISIBILITY.
Imagine that we painted every 12th number, going around a clock. Clearly we'd just be painting the same number every time, because 12/12 = 1, with no remainder.
If you painted every 2 numbers (starting at 12), you'd paint the 6 even numbers, but you'd keep repeating those and never paint the odds, because 12/2 = 6, with no remainder. Any FACTOR of 12 will divide the 12 spaces up evenly, so you'll land back where you started every time. So, we can eliminate A, B, and C.
D and E are trickier, since neither 7 nor 9 is a factor of 12. However, 12 and 9 share a factor of 3, which is why, as Rich demonstrated, the pattern of counting by 9's on a clock face of 12 will land you in the same places: 9, 6, 3, and 12. Think about the multiples of 9 and 12:
9x: 9, 18, 27, 36, 45, 54, 63, 72, 81, 90... --> every 4th multiple of 9 is also a multiple of 12, and the others have remainders of 9, 6, or 3.
12x: 12, 24, 36, 48, 60, 72, 84... --> every 3rd multiple of 12 is also a multiple of 9
Because 7 and 12 don't share any factors, there won't be any particular pattern that they share. Here are the multiples of 7, and the remainders we get when we divide by 12:
7 --> R: 7
14 --> R: 2
21 --> R: 9
28 --> R: 4
35 --> R: 11
42 --> R: 6
49 --> R: 1
56 --> R: 8
63 --> R: 3
70 --> R: 10
77 --> R: 5
84 --> R: 0
We'll eventually hit every remainder up to 11, so we'll hit every number on the clock face.
Ceilidh Erickson
EdM in Mind, Brain, and Education
Harvard Graduate School of Education
