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by bblast » Thu Aug 11, 2011 7:35 am
If p, q, and R are non-negative integers such that the remainder when (10^p) - q is divided by 3 is equal to R, what is the value of R?

p = 7
q = 4

[spoiler]Is the answer B and not C because its given that R is a positive integer ? Because when P=0, R is not equal to 0 by statement 2.[/spoiler]

{edited to add p to the power of 10.}
Last edited by bblast on Thu Aug 11, 2011 8:19 am, edited 1 time in total.
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by gmatboost » Thu Aug 11, 2011 8:05 am
We can use some examples to see that each statement is Insufficient.

1. q = 1 vs q = 2
2. p = 1 vs. p = 2

I am not sure I understand your question though.
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by bblast » Thu Aug 11, 2011 8:16 am
gmatboost wrote:We can use some examples to see that each statement is Insufficient.

1. q = 1 vs q = 2
2. p = 1 vs. p = 2

I am not sure I understand your question though.

Hi, I am sorry, i mistyped the question , please take another look at it.i made the corrections.
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by gmatboost » Thu Aug 11, 2011 8:36 am
Okay, I see what you're saying.

I think that on the actual GMAT the question would give some sort of boundary, like p > 0, in order to avoid cases in which 10^p - q < 0.

Having said that, if q = 4 and p = 0, then 10^p - q = 1 - 4 = -3, which is "divisible by 3." I say that in quotes because we don't really talk about negative numbers as being divisible by numbers, or as being factors or multiples of numbers.

When q = 4 and p > 0, 10^p - q does seem to generate multiples of 3: 6, 96, 996, 9996, and so on. Every number will be a strong of 9's with a 6 at the end. Since 9 and 6 are multiples of 3, the sum of the digits will always be a multiple of 3, which means every such number will be divisible by 3.

So it does look like B is right, I think on the test you wouldn't have to worry about the p = 0 case.
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by GmatKiss » Thu Aug 11, 2011 8:41 am
OA should be B

anything to the power 10 will end up with 0
from 2, 10-7 , 100-7 1000-7 .. will end up with 3 or 93, which is divisible by 3.

1) is not sufficient to make the end value divisible by 3!
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