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Re: are a, b, and c consecutive integers? [#permalink]
22 Oct 2009, 22:59
4
This post received KUDOS
Expert's post
4
This post was BOOKMARKED
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) \(\frac{1}{a}-\frac{1}{b}=\frac{1}{c}\) --> \((b-a)c=ab\) --> \((a+1-a)(a+2)=a(a+1)\) --> \(a^2=2\) --> \(a^2=2\) as \(a\) is positive integer this equation has no positive integer roots --> \(a\), \(b\), \(c\), are not consecutive positive integers. Sufficient.
(2) \(a+c=b^2-1\) --> \(a+a+2=(a+1)^2-1\) --> \(2a+2=a^2+2a+1-1\) --> \(a^2=2\). The same here: \(a\), \(b\), \(c\), are not consecutive positive integers. Sufficient.
Re: If a, b, and c are positive integers, with a < b < c, [#permalink]
24 Oct 2013, 06:05
1
This post received KUDOS
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 co-prime 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)(b-1)
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^2-1
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
Last edited by jlgdr on 24 Oct 2013, 07:05, edited 1 time in total.
Re: If a, b, and c are positive integers, with a < b < c, [#permalink]
24 Oct 2013, 06:41
Expert's post
jlgdr 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 = 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 co-prime 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)(b-1)
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^2-1
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 (E)
Please let me know whether this makes sense. If it does, throw me some Kudos!
Re: If a, b, and c are positive integers, with a < b < c, [#permalink]
24 Oct 2013, 07:06
Bunuel wrote:
jlgdr 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 = 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 co-prime 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)(b-1)
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^2-1
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 (E)
Please let me know whether this makes sense. If it does, throw me some Kudos!
Re: If a, b, and c are positive integers, with a < b < c, [#permalink]
04 Feb 2014, 05:24
I assumed a,b,c to be consecutive and substituted b-1,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.
Re: are a, b, and c consecutive integers? [#permalink]
14 May 2014, 03:40
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) \(\frac{1}{a}-\frac{1}{b}=\frac{1}{c}\) --> \((b-a)c=ab\) --> \((a+1-a)(a+2)=a(a+1)\) --> \(a^2=2\) --> \(a^2=2\) as \(a\) is positive integer this equation has no positive integer roots --> \(a\), \(b\), \(c\), are not consecutive positive integers. Sufficient.
(2) \(a+c=b^2-1\) --> \(a+a+2=(a+1)^2-1\) --> \(2a+2=a^2+2a+1-1\) --> \(a^2=2\). 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 ?
Re: are a, b, and c consecutive integers? [#permalink]
14 May 2014, 05:35
Expert's post
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) \(\frac{1}{a}-\frac{1}{b}=\frac{1}{c}\) --> \((b-a)c=ab\) --> \((a+1-a)(a+2)=a(a+1)\) --> \(a^2=2\) --> \(a^2=2\) as \(a\) is positive integer this equation has no positive integer roots --> \(a\), \(b\), \(c\), are not consecutive positive integers. Sufficient.
(2) \(a+c=b^2-1\) --> \(a+a+2=(a+1)^2-1\) --> \(2a+2=a^2+2a+1-1\) --> \(a^2=2\). 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. _________________
Re: are a, b, and c consecutive integers? [#permalink]
14 May 2014, 06:29
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) \(\frac{1}{a}-\frac{1}{b}=\frac{1}{c}\) --> \((b-a)c=ab\) --> \((a+1-a)(a+2)=a(a+1)\) --> \(a^2=2\) --> \(a^2=2\) as \(a\) is positive integer this equation has no positive integer roots --> \(a\), \(b\), \(c\), are not consecutive positive integers. Sufficient.
(2) \(a+c=b^2-1\) --> \(a+a+2=(a+1)^2-1\) --> \(2a+2=a^2+2a+1-1\) --> \(a^2=2\). 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]
Re: If a, b, and c are positive integers, with a < b < c, [#permalink]
28 Jun 2015, 03:05
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If a, b, and c are positive integers, with a < b < c, [#permalink]
02 Feb 2016, 18:08
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+1-a/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.
gmatclubot
If a, b, and c are positive integers, with a < b < c,
[#permalink]
02 Feb 2016, 18:08
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