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M16 Q35 PS from GMAT club tests [#permalink]
14 Aug 2008, 08:25

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\(X\) , \(A\) , 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) \(A + 2 = B + 1\)

Re: PS from GMAT club test M16 Q35 [#permalink]
14 Aug 2008, 08:42

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durgesh79 wrote:

spent 10 minutes on this one .... and finally guessed it and it was right ...

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

A is even X+B is divisible by A X-1 is divisible by A B = A-1 A+2 = B+1

Re: PS from GMAT club test M16 Q35 [#permalink]
14 Aug 2008, 08:58

14

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durgesh79 wrote:

spent 10 minutes on this one .... and finally guessed it and it was right ...

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

A is even X+B is divisible by A X-1 is divisible by A B = A-1 A+2 = B+1

D firm:

When X is divided by A, the remainder is B. -- means A>B (remainder always less A)

Re: M16 Q35 PS from GMAT club tests [#permalink]
27 Apr 2010, 22:07

Also from substitution method it we can see that by taking numbers for A and B , taking 3 and 2 the closer number would be 5, as the conditions get satisfied so we can assure that 3 and 2 are A,B. Thus satisfies the conditions.

Re: M16 Q35 PS from GMAT club tests [#permalink]
28 Apr 2011, 09:20

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spent more than 10 min trying to solve, finally substituted numbers (A=3, B=2) to arrive at (D) ... great approach by x2suresh ... divisor>remainder ... life can be so simple _________________

Re: M16 Q35 PS from GMAT club tests [#permalink]
02 May 2012, 10:29

Excellent Concept Problem: D is correct answer.

When X is divided by A, the remainder is B. => A > B When X is divided by B, the remainder is (A-2). => B > A - 2 Combining the two inequalities: A > B > A - 2 i.e.: B is between A and (A-2), so B must be equal to A-1. B = A - 1. (D) is the correct answer.

Re: M16 Q35 PS from GMAT club tests [#permalink]
03 May 2013, 04:18

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Expert's post

durgesh79 wrote:

\(X\) , \(A\) , 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) \(A + 2 = B + 1\)

spent 10 minutes on this one .... and finally guessed it and it was right ...

BELOW IS REVISED VERSION OF THIS QUESTION:

If \(x\), \(a\), and \(b\) are positive integers such that when \(x\) is divided by \(a\), the remainder is \(b\) and when \(x\) is divided by \(b\), the remainder is \(a-2\), then 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 \(b=a-1\)).

Re: M16 Q35 PS from GMAT club tests [#permalink]
18 Apr 2014, 05:50

Bunuel wrote:

durgesh79 wrote:

\(X\) , \(A\) , 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) \(A + 2 = B + 1\)

spent 10 minutes on this one .... and finally guessed it and it was right ...

BELOW IS REVISED VERSION OF THIS QUESTION:

If \(x\), \(a\), and \(b\) are positive integers such that when \(x\) is divided by \(a\), the remainder is \(b\) and when \(x\) is divided by \(b\), the remainder is \(a-2\), then 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 \(b=a-1\)).

Answer: D.

Do you have any suggestions on how one can be quick at finding this insight? (And finding the correct insight quickly in general?)