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If a, b, and c are positive distinct integers, is (a/b)/c an

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If a, b, and c are positive distinct integers, is (a/b)/c an [#permalink] New post 13 Jul 2010, 08:48
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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
[Reveal] Spoiler: OA

Last edited by Bunuel on 08 Nov 2012, 02:18, edited 1 time in total.
Edited the question.
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Re: m05 [#permalink] New post 13 Jul 2010, 09:04
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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Re: m05 [#permalink] New post 13 Jul 2010, 09:29
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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] New post 07 Nov 2012, 19:37
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] New post 08 Nov 2012, 02:27
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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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Re: m05 [#permalink] New post 22 Oct 2013, 19:52
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!
Re: m05   [#permalink] 22 Oct 2013, 19:52
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