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If a, b, and c are positive distinct integers, is (a/b)/c an [#permalink]
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If a, b, and c are positive distinct integers, is (a/b)/c an integer? (1) c = 2 (2) a = b+c
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Last edited by Bunuel on 08 Nov 2012, 03:18, edited 1 time in total.
Edited the question.



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A humble request: I think there is a separate section for GMAT Club tests. There you can find the already discussed questions too. Now coming to the problem. 1 is clearly insufficient. 2 is sufficient. No combination can be found in which \((a/b)/c\) is an integer provided \(a= b + c\) & \(a, b, c\) are distinct.
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bibha wrote: If a, b, and c are positive distinct integers, is (a/b)/c an integer? 1. c=2 2. a = b+c First of all \(\frac{\frac{a}{b}}{c}=\frac{a}{bc}\). So the question becomes is \(\frac{a}{bc}=integer\) true? (1) \(c=2\) > is \(\frac{a}{2b}=integer\). Clearly insufficient. If \(a=1\), then answer is NO, but if \(a=4\) and \(b=1\), then the answer is YES. (2) \(a=b+c\) > \(\frac{a}{bc}=\frac{b+c}{bc}=\frac{b}{bc}+\frac{c}{bc}=\frac{1}{c}+\frac{1}{b}\). As \(b\) and \(c\) are distinct integers then \(\frac{1}{c}+\frac{1}{b}\) won't be an integer. Sufficient. (Side note: if \(b\) and \(c\) were not distinct integers then \(\frac{1}{c}+\frac{1}{b}\) could be an integer in the following cases: \(b=c=1\) and \(b=c=2\)). Answer: B. Hope it helps.
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Re: If a, b, and c are positive distinct integers, is (a/b)/c an [#permalink]
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07 Nov 2012, 20:37
Bunuel,
Stupid question. But....
When I see the question: Is a/bc = Integer. My mind immediately multiplies the equation by bc on both sides and comes up with a new question...
Is a = (integer) (bc)
since both b and c are distinct integers and thus bc will be an integer, the question is asking if a = integer.
If this is accurate, then isn't option B a straightforward Yes? If A = B +C, then A is just an addition of two distinct integers, making A itself an integer, answering the original question.
Can you help point out where I went wrong?
Thanks.



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Re: If a, b, and c are positive distinct integers, is (a/b)/c an [#permalink]
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08 Nov 2012, 03:27
navalpike wrote: Bunuel,
Stupid question. But....
When I see the question: Is a/bc = Integer. My mind immediately multiplies the equation by bc on both sides and comes up with a new question...
Is a = (integer) (bc)
since both b and c are distinct integers and thus bc will be an integer, the question is asking if a = integer.
If this is accurate, then isn't option B a straightforward Yes? If A = B +C, then A is just an addition of two distinct integers, making A itself an integer, answering the original question.
Can you help point out where I went wrong?
Thanks. If the question could be rephrased as "is \(a\) an integer", then we wouldn't need any statement to answer the question, since the stem directly says that \(a\) is an integer. The question asks whether \(\frac{a}{bc}\) is an integer or whether \(a\) is a multiple of \(bc\). Hope it helps.
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Bunuel wrote: bibha wrote: If a, b, and c are positive distinct integers, is (a/b)/c an integer? 1. c=2 2. a = b+c First of all \(\frac{\frac{a}{b}}{c}=\frac{a}{bc}\). So the question becomes is \(\frac{a}{bc}=integer\) true? (1) \(c=2\) > is \(\frac{a}{2b}=integer\). Clearly insufficient. If \(a=1\), then answer is NO, but if \(a=4\) and \(b=1\), then the answer is YES. (2) \(a=b+c\) > \(\frac{a}{bc}=\frac{b+c}{bc}=\frac{b}{bc}+\frac{c}{bc}=\frac{1}{c}+\frac{1}{b}\). As \(b\) and \(c\) are distinct integers then \(\frac{1}{c}+\frac{1}{b}\) won't be an integer. Sufficient. (Side note: if \(b\) and \(c\) were not distinct integers then \(\frac{1}{c}+\frac{1}{b}\) could be an integer in the following cases: \(b=c=1\) and \(b=c=2\)). Answer: B. Hope it helps. Thanks for the response. I also chose B but took over 3 minutes because I kept checking to see whether there might be a situation in which the reciprocals of 2 distinct integers can add up to produce an integer. Wondering if there is a mathematical concept / theory that can prove this will never be the case? I know it's unnecessary but will help strengthen our understanding of how reciprocals of positive integers function. One thing that I can think of: 1.Imagine a number line with 0, 1, 2 and so on. Now both the reciprocals of positive integers a and b will lie between 0 and 1 (inclusive)  no other form is possible given that a and b are positive integers (don't think about distinct integers just yet) 2. Now if the sum of these two fractions (1/a and 1/b) must be an integer there are two possibilities  either a. they sum up to 1 or b. they sum up to 2 The MAXIMUM value of 1/a or 1/b is 1  so the maximum sum is 1+1 or 2 (to maximize 1/a we must minimize a, which is a positive integer and 1 is the smallest positive integer) 3. The question now becomes can 1/a+1/b be 1 or 2 with a and be being DIFFERENT integers a. Sum of 1  this is only possible at 1/2 and 1/2 as no other fraction can be written of the form 1/integer and summed up to add 1 > my question to the group is  can this be theorized? b. Sum of 2  this is only possible at 1 and 1 as no 2 fractions between 0 and 1 can be summed up to give you 2 (test extreme case 0.999+0.998 = 1.997 which is less than 2) B produces a definite answer  therefore B Thoughts welcome!



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