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If x, a, and b are positive integers such that when x is
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29 Sep 2010, 22:27
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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 a2, then which of the following must be true? A. a is even B. x+b is divisible by a C. x1 is divisible by a D. b=a1 E. a+2=b+1
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Re: m16 #35
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29 Sep 2010, 22:49
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 \(a2\), then which of the following must be true?A. \(a\) is even B. \(x+b\) is divisible by \(a\) C. \(x1\) is divisible by \(a\) D. \(b=a1\) 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 \(a2\) > \(x=bp+(a2)\) > \(remainder=(a2)<b=divisor\). So we have that: \(a2<b<a\), as \(a\) and \(b\) are integers, then it must be true that \(b=a1\) (there is only one integer between \(a2\) and \(a\), which is \(a1\) and we are told that this integer is \(b\), hence \(b=a1\)). Answer: D.
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Re: m16 #35
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29 Sep 2010, 23:00
Awesome Bunuel. Was trying hard to figure out the answer. All the information was just readily available but was looking all around. +1 to you.
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Re: m16 #35
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29 Sep 2010, 23:06
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..?



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Re: m16 #35
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29 Sep 2010, 23:18
Simply amazing bunuel...... you are a genius....
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Re: m16 #35
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29 Sep 2010, 23:29
cool, BUenel, i had go arround with the formulations but can not find that A2<B tks much



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Re: m16 #35
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29 Sep 2010, 23:39



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Re: m16 #35
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30 Sep 2010, 05:57
Bunuel wrote: Priceless info.Ton thanks



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Re: m16 #35
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01 Oct 2010, 07:29
Awesome solution Bunuel. +1 Great question Kudos to GMAT club tests



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Re: m16 #35
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01 Oct 2010, 10:35
Awesome!!! I didnt get the fastest way, but through number plugin I was able to figure out the solution. But your solution is so much better!!



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01 Oct 2010, 19:49
nice and efficient !
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02 Oct 2010, 20:02
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!



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20 Oct 2010, 04:38
Nice approach. +1 kudos



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Re: If x, a, and b are positive integers such that when x is div
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04 Sep 2012, 01:14
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 A2, 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
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05 Sep 2012, 23:56
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 \(a2\), then which of the following must be true?
A. \(a\) is even B. \(x+b\) is divisible by \(a\) C. \(x1\) is divisible by \(a\) D. \(b=a1\) 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 \(a2\) > \(x=bp+(a2)\) > \(remainder=(a2)<b=divisor\).
So we have that: \(a2<b<a\), as \(a\) and \(b\) are integers, then it must be true that \(b=a1\) (there is only one integer between \(a2\) and \(a\), which is \(a1\) and we are told that this integer is \(b\), hence \(b=a1\)).
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
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20 Apr 2013, 12:57
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
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19 Jun 2013, 04:52



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Re: If x, a, and b are positive integers such that when x is
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26 Sep 2013, 20:26
Way bunuel approch the problem is fantastic. I dont know whether i would be able to think the way he does . . tat too in time pressure +1 kudos



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Re: If x, a, and b are positive integers such that when x is
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03 Nov 2013, 01:50
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 a2... > 5/2=2+1/2
Both work. If we plug numbers into the answers, only D will work.
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Re: m16 #35
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06 Nov 2013, 14:10
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 \(a2\), then which of the following must be true?
A. \(a\) is even B. \(x+b\) is divisible by \(a\) C. \(x1\) is divisible by \(a\) D. \(b=a1\) 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 \(a2\) > \(x=bp+(a2)\) > \(remainder=(a2)<b=divisor\).
So we have that: \(a2<b<a\), as \(a\) and \(b\) are integers, then it must be true that \(b=a1\) (there is only one integer between \(a2\) and \(a\), which is \(a1\) and we are told that this integer is \(b\), hence \(b=a1\)).
Answer: D. Hi Bunuel, I solved this problem with a bit different approach x = p*a + b..........eqn(1) and x = q*b + (a2)..............eqn(2) now, equating eqn(1) and eqn(2) p*a + b = q*b + (a2) a(p1) = b(q1)  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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