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M05-36

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M05-36  [#permalink]

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New post 15 Sep 2014, 23:26
10
00:00
A
B
C
D
E

Difficulty:

  95% (hard)

Question Stats:

40% (01:12) correct 60% (01:27) wrong based on 181 sessions

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Re M05-36  [#permalink]

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New post 15 Sep 2014, 23:26
Official Solution:


First of all: \(\frac{(\frac{a}{b})}{c}=\frac{a}{bc}\).

(1) \(\frac{a}{c} = 3\). Substitute the value of \(\frac{a}{c}\) in the equation given above: \(\frac{a}{bc}=\frac{3}{b}\). Now, if \(b\) is 1 or 3 then the answer is YES but if \(b\) is some other positive integer then the answer is NO. Not sufficient.

(2) \(a = b + c\). Substitute the value of \(a\) in the equation given above: \(\frac{a}{bc}=\frac{b+c}{bc}=\frac{b}{bc}+\frac{c}{bc}=\frac{1}{c}+\frac{1}{b}\). Now, this expression to be an integer (1) either \(b=c=1\) must be true of \(b=c=2\) must be true, but we are told that given unknowns are distinct, so neither of option is possible. So, \(\frac{1}{c}+\frac{1}{b} \ne \text{integer}\). Sufficient.


Answer: B
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Re M05-36  [#permalink]

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New post 12 Aug 2015, 05:22
what is b=-1 and c=1 i n that case also 1/b+ 1/c= 0 (which is integer) fulfills the condition. So answer should be E
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Re: M05-36  [#permalink]

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New post 12 Aug 2015, 10:13
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fearose wrote:
what is b=-1 and c=1 i n that case also 1/b+ 1/c= 0 (which is integer) fulfills the condition. So answer should be E


b= -1 is not possible since the question says "If a, b, and c are distinct positive integers".
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M05-36  [#permalink]

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New post 09 Oct 2015, 14:53
Bunuel wrote:
If \(a\), \(b\), and \(c\) are distinct positive integers, is \(\frac{(\frac{a}{b})}{c}\) an integer?


(1) \(\frac{a}{c} = 3\)

(2) \(a = b + c\)


Simplifying expression \(\frac{(\frac{a}{b})}{c}\) gives a/bc

Now

Statement 1 : a/c=3

therefore a/bc=3/b, No data related to b is given.

Insufficient

Statement 2 : a=b+c

a/bc=(b+c)/bc=1/b+1/c......(i)

As b and c are distinct. Therefore expression will never be an integer.

Sufficient.

Answer : B
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Re: M05-36  [#permalink]

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New post 30 Oct 2015, 01:55
Statement 1 : sufficient data is not provided.

Statement 2 : (a/b)/c = a/ (b *c)

Now, a = b+c and a,b,c are distinct positive integers,hence ( b*c ) will always be greater than a. Hence result wont be an integer
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Re M05-36  [#permalink]

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New post 20 Jun 2016, 15:39
I think this is a high-quality question and I agree with explanation.
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Re: M05-36  [#permalink]

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New post 22 Oct 2016, 13:10
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Bunuel,

How did you arrive at this? - "Now, this expression to be an integer (1) either b=c=1 must be true or b=c=2 must be true". Why cant b and c be two distinct positive integers such that their decimal add upto 1? How do you concretely say b=c=1 or b=c=2. How do you rule out other possibilities.

I thought ( even though it may not be possible ), we may have numbers like 1/b = 0.6 & 1/c = 0.4 which add upto 1. I thought we may have such decimals which add upto 1. How do you rule out every such possibility?
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Re M05-36  [#permalink]

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New post 01 Jan 2017, 07:03
I think this is a high-quality question and I agree with explanation.
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Re: M05-36  [#permalink]

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New post 03 Jul 2017, 06:16
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vrgmat wrote:
Bunuel,

How did you arrive at this? - "Now, this expression to be an integer (1) either b=c=1 must be true or b=c=2 must be true". Why cant b and c be two distinct positive integers such that their decimal add upto 1? How do you concretely say b=c=1 or b=c=2. How do you rule out other possibilities.

I thought ( even though it may not be possible ), we may have numbers like 1/b = 0.6 & 1/c = 0.4 which add upto 1. I thought we may have such decimals which add upto 1. How do you rule out every such possibility?


Even I was thinking on the same lines but later on, I tried to figure it out on my own.
Addressing your query:
Two positive fractions will add up to 1, only when they are in this format : 1/x + (x-1)/x i.e. 1/3 + 2/3 or 1/10 + 9/10.
And x & x-1 are co-prime i.e. there is no common factors between them hence they can never be simplified further.
So, if one of the denominator is Integer, other denominator has to be non-integer to add up to 1.
Or, if both the denominators are integer, sum will never be equal to 1.
Hence, for Statement (2), equation : a/bc= 1/c + 1/b will never be Integer if both c & b will be integer.
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Re M05-36  [#permalink]

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New post 14 Jul 2018, 01:20
I think this is a high-quality question and I agree with explanation.
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Re M05-36 &nbs [#permalink] 14 Jul 2018, 01:20
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