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Problem Solving — algebra and arithmetic (GMAT Focus Edition)
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by sunsun2z16 » Sun Oct 16, 2016 6:41 am
13. The 12 numbers on a circular clock are equally spaced around the edges of the clock. Belinda chooses an integer, n, that is greater than 1. Beginning at a randomly chosen number, Belinda goes around the circle counterclockwise and paints in every nth number. She continues going around and around the clock, painting in every nth number, until all twelve numbers on the clock are painted. Which of the following could have been Belinda's integer n?
A. 2
B. 3
C. 6
D. 7
E. 9

I have this question but i don't really understand how to solve it, its answer is D.7
please explain the answer for me...Tks so much
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by [email protected] » Sun Oct 16, 2016 9:41 am
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

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by ceilidh.erickson » Mon Oct 31, 2016 7:03 am
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.
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by Matt@VeritasPrep » Thu May 11, 2017 7:56 pm
How I'd think about it:

Each number on the clock has some remainder when n. To get to every number on the clock, you need to be able to hit each remainder. So if your number ever repeats before you've gotten through all 12 numbers, you're in big trouble: the pattern will just restart the same cycle! (For instance, say n = 2, and we start at 12:00. We go 12:00 -> 10:00 -> 8:00 -> 6:00 -> 4:00 -> 2:00 -> 12:00. Now it'll just repeat!)

With that in mind, you need a number that won't repeat too quickly, and it should be clear that 2, 3, and 6, as factors of 12 -- the 12 numbers on the clockface -- will do so. 9 is also suspicious, since it shares a factor of 3 with 12, so let's test it to see. Suppose n = 9, and we start at 12:00. 12:00 -> 3:00 -> 6:00 -> 9:00 -> 12:00 ... there it is! So that won't work either.

By process of elimination, then, the answer is 7. The longer, more technical answer is that 7 works because it shares no prime factors with 12, but under test conditions I think the approach above is better.

Would add that if I saw this on a test and had no idea what to do, I'd immediately start grinding out the answer choices. They don't take that long to evaluate if you work quickly, possibly under two minutes, and you don't really need to know what's going on: you can let the numbers tell you.
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