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If a, b, and c are positive integers, with a < b < c, [#permalink]
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Updated on: 10 Jun 2012, 23:02
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If a, b, and c are positive integers, with a < b < c, are a, b, and c consecutive integers? (1) 1/a – 1/b = 1/c (2) a + c = b^2 – 1
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Originally posted by SMAbbas on 22 Oct 2009, 23:27.
Last edited by Bunuel on 10 Jun 2012, 23:02, edited 1 time in total.
Edited the question.



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Re: are a, b, and c consecutive integers? [#permalink]
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SMAbbas wrote: If a, b, and c are positive integers, with a < b < c, are a, b, and c consecutive integers? (1) 1/a – 1/b = 1/c (2) a + c = b2 – 1 IMO D, Please verify my answer If \(a\), \(b\), \(c\) are consecutive and \(a<b<c\), then must be true that: \(b=a+1\) and \(c=a+2\). (1) \(\frac{1}{a}\frac{1}{b}=\frac{1}{c}\) > \((ba)c=ab\) > \((a+1a)(a+2)=a(a+1)\) > \(a^2=2\) > \(a^2=2\) as \(a\) is positive integer this equation has no positive integer roots > \(a\), \(b\), \(c\), are not consecutive positive integers. Sufficient. (2) \(a+c=b^21\) > \(a+a+2=(a+1)^21\) > \(2a+2=a^2+2a+11\) > \(a^2=2\). The same here: \(a\), \(b\), \(c\), are not consecutive positive integers. Sufficient. Answer D.
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Re: are a, b, and c consecutive integers? [#permalink]
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27 Oct 2009, 07:01
I think I must have misunderstood the basic rule on DS
Like this question, when we find it SUFFICIENT to say "it can't be a real number," it should be good enought to choose (D).
I thought I could choose (D) as long as I could verify YES for quesitions.



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Re: are a, b, and c consecutive integers? [#permalink]
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22 May 2010, 08:19
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i approached the question differently ... this may not be the best way but i'm going to share it anyway ... sharing is caring
a, b, c > 0 [positive integers] b=a+1, c=a+2 [a, b, c are consecutive integers]
(1) 1/a  1/b = 1/c 1/a  1/a+1 = 1/a+2 1/a(a+1) = 1/(a+2) a+2 = a(a+1) c = a*b [c=a+2, b=a+1]
if a, b, c are consecutive ... the above equation in bold can never be true 2 ? 0*1 3 ? 1*2 4 ? 2*3
1 is sufficient to explain that a, b, c are not consecutive integers
(2) a+c = (b1)(b+1) a+c = a*c [a=b1, c=b+1]
if a, b, c are consecutive ... the above equation in bold can never be true 0+2 ? 0*2 1+3 ? 1*3 2+4 ? 2*4
2 is sufficient to explain that a, b, c are not consecutive integers
ans is D
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Re: are a, b, and c consecutive integers? [#permalink]
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22 May 2010, 16:01
i approached the question differently ... this may not be the best way but i'm going to share it anyway ... sharing is caring As long as you get the right answer, any approach should be fine !



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Re: If a, b, and c are positive integers, with a < b < c, [#permalink]
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SMAbbas wrote: If a, b, and c are positive integers, with a < b < c, are a, b, and c consecutive integers?
(1) 1/a – 1/b = 1/c
(2) a + c = b^2 – 1 Hey guys  The way I did it was (1) 1/a – 1/b = 1/c > 1/a = 1/b + 1/c LCM of 'b' and 'c' would never be smaller than b or c because these numbers are coprime which means that they do not share any common integer factor other than 1. Now since c>b>a then they following the logic above they can't be consecutive integers. Sufficient (2) a + c = b^2 – 1 > a+c = (b+1)(b1) Now if a,b,c were consecutive integers it will be true that (a+c)/2 is equal to b So we would have that a+c = 2b Now if we replace a+c = 2b in the first equation we would have that 2b = b^21 And when trying to solve this quadratic we would realize that b is not an integer, since we would have to use the formula So it contradicts the information that a,b,c are integers. Hence a,b,c cannot be consecutive integers Insuff Answer (D) Please let me know whether this makes sense. If it does, throw me some Kudos! Cheers J
Originally posted by jlgdr on 24 Oct 2013, 07:05.
Last edited by jlgdr on 24 Oct 2013, 08:05, edited 1 time in total.



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Re: If a, b, and c are positive integers, with a < b < c, [#permalink]
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24 Oct 2013, 21:53
1) 1/a – 1/b = 1/c let a=1 b=2 c=3 1/21/3 doesnt equal to 1/4 (2) a + c = b^2 – 1 let a(even) b(odd) c(even) a + c = b^2 – 1 = even +even =odd^1=even  ok! let a(odd) b(even) c(odd) a + c = b^2 – 1 = odd +odd doesnt equal to even^1 not ok!
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Re: If a, b, and c are positive integers, with a < b < c, [#permalink]
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04 Feb 2014, 06:24
I assumed a,b,c to be consecutive and substituted b1,b,b+1 in each statement independently. (1) gives b^2=2b+1 and b is not an integer hence my assumption that a,b,c are consecutive cannot be true. sufficient.(2) also gives b^2 = 2b+1. same as (1) hence sufficient.+1 D. Hope it made sense.
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Re: are a, b, and c consecutive integers? [#permalink]
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14 May 2014, 04:40
Bunuel wrote: SMAbbas wrote: If a, b, and c are positive integers, with a < b < c, are a, b, and c consecutive integers? (1) 1/a – 1/b = 1/c (2) a + c = b2 – 1 IMO D, Please verify my answer If \(a\), \(b\), \(c\) are consecutive and \(a<b<c\), then must be true that: \(b=a+1\) and \(c=a+2\). (1) \(\frac{1}{a}\frac{1}{b}=\frac{1}{c}\) > \((ba)c=ab\) > \((a+1a)(a+2)=a(a+1)\) > \(a^2=2\) > \(a^2=2\) as \(a\) is positive integer this equation has no positive integer roots > \(a\), \(b\), \(c\), are not consecutive positive integers. Sufficient. (2) \(a+c=b^21\) > \(a+a+2=(a+1)^21\) > \(2a+2=a^2+2a+11\) > \(a^2=2\). The same here: \(a\), \(b\), \(c\), are not consecutive positive integers. Sufficient. Answer D. Hi Bunuel, Thanks for the answer. One question  what is the meaning of "this eq has no positive integer root ?". It is because the eq has A = + 2 ?



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Re: are a, b, and c consecutive integers? [#permalink]
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14 May 2014, 06:35
gauravsoni wrote: Bunuel wrote: SMAbbas wrote: If a, b, and c are positive integers, with a < b < c, are a, b, and c consecutive integers? (1) 1/a – 1/b = 1/c (2) a + c = b2 – 1 IMO D, Please verify my answer If \(a\), \(b\), \(c\) are consecutive and \(a<b<c\), then must be true that: \(b=a+1\) and \(c=a+2\). (1) \(\frac{1}{a}\frac{1}{b}=\frac{1}{c}\) > \((ba)c=ab\) > \((a+1a)(a+2)=a(a+1)\) > \(a^2=2\) > \(a^2=2\) as \(a\) is positive integer this equation has no positive integer roots > \(a\), \(b\), \(c\), are not consecutive positive integers. Sufficient. (2) \(a+c=b^21\) > \(a+a+2=(a+1)^21\) > \(2a+2=a^2+2a+11\) > \(a^2=2\). The same here: \(a\), \(b\), \(c\), are not consecutive positive integers. Sufficient. Answer D. Hi Bunuel, Thanks for the answer. One question  what is the meaning of "this eq has no positive integer root ?". It is because the eq has A = + 2 ? We are given that \(a\) is a positive integer, while from \(a^2=2\), \(a=\sqrt{2}\) or \(a=\sqrt{2}\). Hence our assumption that a, b, and c were consecutive integers was wrong.
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Re: are a, b, and c consecutive integers? [#permalink]
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14 May 2014, 07:29
gauravsoni wrote: Bunuel wrote: SMAbbas wrote: If a, b, and c are positive integers, with a < b < c, are a, b, and c consecutive integers? (1) 1/a – 1/b = 1/c (2) a + c = b2 – 1 IMO D, Please verify my answer If \(a\), \(b\), \(c\) are consecutive and \(a<b<c\), then must be true that: \(b=a+1\) and \(c=a+2\). (1) \(\frac{1}{a}\frac{1}{b}=\frac{1}{c}\) > \((ba)c=ab\) > \((a+1a)(a+2)=a(a+1)\) > \(a^2=2\) > \(a^2=2\) as \(a\) is positive integer this equation has no positive integer roots > \(a\), \(b\), \(c\), are not consecutive positive integers. Sufficient. (2) \(a+c=b^21\) > \(a+a+2=(a+1)^21\) > \(2a+2=a^2+2a+11\) > \(a^2=2\). The same here: \(a\), \(b\), \(c\), are not consecutive positive integers. Sufficient. Answer D. Hi Bunuel, Thanks for the answer. One question  what is the meaning of "this eq has no positive integer root ?". It is because the eq has A = + 2 ? We are given that \(a\) is a positive integer, while from \(a^2=2\), \(a=\sqrt{2}\) or \(a=\sqrt{2}\). Hence our assumption that a, b, and c were consecutive integers was wrong.[/quote] Got it , thanks...



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Re: If a, b, and c are positive integers, with a < b < c, [#permalink]
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15 May 2014, 03:32
Bunuel,
We can also do it by taking a, b and c as (x1), x, (x+1) right? It looks the same to me as taking b = a+1 and c = a+2



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Re: If a, b, and c are positive integers, with a < b < c, [#permalink]
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15 May 2014, 03:42



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Re: If a, b, and c are positive integers, with a < b < c, [#permalink]
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15 May 2014, 03:46
Thanks a lot Bunuel



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If a, b, and c are positive integers, with a < b < c, [#permalink]
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02 Feb 2016, 19:08
took me 2 mins to crack this one.. if integers are consecutive, then we can write as: a a+1 a+2
we have: 1/a  1/a+1 = a+1a/a(a+1) = 1/a^2+1 a^2+1 if a=1, then c=2, impossible. if a=2, then c=5, again impossible. we can see that neither of the options works here.
2. a+c=b^2 +1 a+a+2 = (a+1)^2 +1 2a+2 = a^2+2a a^2=2 this is impossible, since all the numbers are positive, and a must be > than 0 and an integer. we can see that we have a definite answer, and the answer is D.



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Re: If a, b, and c are positive integers, with a < b < c, [#permalink]
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