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If x, a, and b are positive integers such that when x is [#permalink]

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29 Sep 2010, 21:27

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49% (03:10) correct
51% (02:20) wrong based on 611 sessions

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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

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 \(b=a-1\)).

Thanks Bunnel!! Do you suggest using a particular strategy for these problems or using different strategy for every problem and whichever fits the bill for the given question..?

Thanks Bunnel!! Do you suggest using a particular strategy for these problems or using different strategy for every problem and whichever fits the bill for the given question..?

What do you mean by "these problems"? Remainder problems or must be true problems?

Thanks Bunnel!! Do you suggest using a particular strategy for these problems or using different strategy for every problem and whichever fits the bill for the given question..?

What do you mean by "these problems"? Remainder problems or must be true problems?

Bunnel, thanks so much for the compilation! By 'these problems' I meant 'Must be true' questions in which at times you have more than 1 correct answers. My apologies for the lack of clarity their. Your compilation should be enough to practice. Thanks again!

Re: If x, a, and b are positive integers such that when x is div [#permalink]

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04 Sep 2012, 00:14

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sanjoo wrote:

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 divided by A, remainder is B, this implies A > B When divided by B, remainder is A-2, this implies B > A -2

Combining both, B < A < (B + 2) Since, A and B are integers, A = B + 1

Answer is (D) . Cheers!
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Re: If x, a, and b are positive integers such that when x is [#permalink]

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05 Sep 2012, 22:56

Bunuel wrote:

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.

Indeed very nice explanation, but for me, for the person who is not that strong in quants sometimes difficult to keep all that concepts in my head and i am jumping to different approaches. Whenever i see must be true questions i plug in some numbers and see which answer works, since it is must be true questions any numbers should work equally. For example in this problem: lets says x=5, a=3 then b=2, so check all the answers and we see that only d works, but if there will be two answers that work try different numbers till we get only one. It could be time consuming, but when we are asked simple expressions it is easy to find numbers that work well.

Bunuel, do you think there are any pitfalls that i should be aware of?
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Re: If x, a, and b are positive integers such that when x is [#permalink]

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20 Apr 2013, 11:57

My approach of plug in numbers was certainly not the best approach, in a zest of GMAT I dont know why I am loosing to think simple....this was a simple algebra which I complicated with numbers
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Re: If x, a, and b are positive integers such that when x is [#permalink]

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03 Nov 2013, 00:50

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This problem can be solved by plugging numbers. For instance if x=5, a=3, b=2 we get: ...when x is divided by a, the remainder is b... > 5/3=1+2/3 ...and when x is divided by b, the remainder is a-2... > 5/2=2+1/2

Both work. If we plug numbers into the answers, only D will work.

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.

Hi Bunuel,

I solved this problem with a bit different approach x = p*a + b..........eqn(1) and x = q*b + (a-2)..............eqn(2)

now, equating eqn(1) and eqn(2) p*a + b = q*b + (a-2)

a(p-1) = b(q-1) - 2 if we put p = q = 3

we get, 2a = 2b - 2 or a = b - 1 or a + 2 = b + 1 which is option E

would pl tell me where am i wrong with my approach?? Thanks.

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