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# very good one

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14 Jun 2010, 14:00
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I dont usually post on Quant Forum, but I consider this an interesting one:

X, A, and B are positive integers. When X is divided by A , the remainder is B . If when X is divided by B , the remainder is A-2 , which of the following must be true?

a)A is even
b)X+B is divisible by A
c)X-1 is divisible by A
d)B=A-1
e)A+2=B+1
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Last edited by noboru on 14 Jun 2010, 14:50, edited 1 time in total.
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14 Jun 2010, 14:14
noboru wrote:
I dont usually post on Quant Forum, but I consider this an interesting one:

X, A, and B are positive integers. When X is divided by A , the remainder is B . If when X is divided by B , the remainder is A-2 , which of the following must be true?

a)A is even
b)X+B is divisible by A
c)X-1 is divisible by A
d)B=A+1
e)A+2=B+1

When X is divided by A , the remainder is B
=> A>B as the remainder is always left when it is less than the divisor else it is divided again.

when X is divided by B , the remainder is A-2
=> B > A-2 , same explanation.

combine both..... A>B>A-2 => A+2 > B+2 > A

This is only possible when B+1 = A as they are integers.
Take eg of A = 5, B= 4 and A-2 = 3

But this option is not listed , I guess there is a typo error in d or e as they both are equal.
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14 Jun 2010, 14:36
noboru wrote:
I dont usually post on Quant Forum, but I consider this an interesting one:

X, A, and B are positive integers. When X is divided by A , the remainder is B . If when X is divided by B , the remainder is A-2 , which of the following must be true?

a)A is even
b)X+B is divisible by A
c)X-1 is divisible by A
d)B=A+1
e)A+2=B+1

When X is divided by A, the remainder is B --> $$x=aq+b$$ --> $$remainder=b<a=divisor$$ (remainder must be less than divisor);
When X is divided by B, the remainder is A-2 --> $$x=bp+(a-2)$$ --> $$remainder=(a-2)<b=divisor$$.

So we have that: $$a-2<b<a$$, as $$a$$ and $$b$$ are integers, then it must be true that $$b=a-1$$ (there is only one integer between $$a-2$$ and $$a$$, which is $$a-1$$ and we are told that this integer is $$b$$, hence $$b=a-1$$).

No one seems to be necessarily true from answer choices. One thing though D and E are the same, so there must be a typo: if D were $$b=a-1$$, instead of $$b=a+1$$, then D would be the correct answer OR if E were $$a-2=b-1$$, instead of $$a+2=b+1$$, then E would be the correct answer.

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Re: very good one   [#permalink] 14 Jun 2010, 14:36
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