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Remainder

Expert replies
by andrew.ng » Sun Sep 27, 2009 4:02 am
If {n} denote the remainder when 3n is divided by 2 then which of the following is equal to 1 for all positive integers n?

I/ {2n+1}
II/ {2n}+1
III/ 2{n+1}


A/ I only
B/ II only
C/ I and II
D/ III only
E/ II and III

I don't really understand the question. Could somebody help?


Thanks
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Source: — Problem Solving |

imo

by xcusemeplz2009 » Tue Sep 29, 2009 10:31 am
imo c
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by ssmiles08 » Tue Sep 29, 2009 4:53 pm
IMO C too. The question seems a bit weird to me though.
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Re: Remainder

by Stuart@KaplanGMAT » Wed Sep 30, 2009 12:08 pm
andrew.ng wrote:If {n} denote the remainder when 3n is divided by 2 then which of the following is equal to 1 for all positive integers n?

I/ {2n+1}
II/ {2n}+1
III/ 2{n+1}


A/ I only
B/ II only
C/ I and II
D/ III only
E/ II and III

I don't really understand the question. Could somebody help?


Thanks
As with most "wacky definition" questions, we need to start by really decoding the question stem.

Here, we're given this new operation {}. We're told that {n} equals the remainder when 3n is divided by 2. It's often helpful to do a couple of quick concrete examples to help you understand the operation.

So:

{2} = rem of (3*2/2) = rem of 6/2 = 0
{3} = rem of (3*3/2) = rem of 9/2 = 1

We can quickly determine that if n is odd, the remainder will be 1; if n is even, the remainder will be 0. Accordingly:

{odd} = 1
{even} = 0

Now to the exact question: which of the following is equal to 1 for all positive integers n?

We can quickly eliminate (III), since 2*(integer) is never going to equal 1 (even if we didn't understand the operation, we should be able to cross out III). III isn't part of the solution: eliminate (D) and (E).

The other two statements occur with equal frequency in A/B/C, so let's start with I, which looks a bit simpler:

I {2n+1}

Well, we know that 2n will be even, so 2n+1 will be odd. Based on our previous work, we know that {odd}=1, so (I) satisfies the question: eliminate B (only A and C left!).

II {2n} + 1

Again, we know that 2n will be even and that {even}=0. So, II is really:

0 + 1 = 1

and, consequently, II also satisfies the question.

I and II are both always equal to 1: choose C.
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