Hi sunsun2z16,

This question has a visual component to it, so you would likely find it helpful to physically draw a clock so that you can see what numbers would be painted when you consider each of the answer choices. In this way, you really just need to do a bit of 'brute force' to figure out the correct answer. Only one of them will end with all 12 numbers painted - the other 4 will include a repeating pattern.

For example - Answer A; counter-clockwise, if every 2nd number gets painted, then you'll go around the clock painting every other number again and again.

eg. Start at 12 and paint every 2nd number.... you would paint 10, 8, 6, 4, 2, 12, 10, 8, 6, 4, 2, etc. but you wouldn't paint any of the odd numbers. Thus, this cannot be the answer.

In that same way, there are a couple of other obvious answers to eliminate: B and C. Going counter-clockwise and painting every 3rd number or every 6th number would lead to a repeated sequence of numbers.

Answer E takes a little more work (because you have to go around the clock several times to see the pattern):

eg. Start at 12 and paint every 9th number counter-clockwise... you would paint 3, 6, 9, 12, 3, 6, 9, 12.

Since all 4 of those options leads to a repeating pattern that does not hit all of the numbers, we're left with just one answer (and it must be the correct one).

Final Answer: D

GMAT assassins aren't born, they're made,
Rich

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.