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# If a, b, and c are positive integers, with a < b < c,

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

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22 Oct 2009, 23:27
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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
[Reveal] Spoiler: OA

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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22 Oct 2009, 23:59
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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,

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.

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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 = (b-1)(b+1)
a+c = a*c [a=b-1, 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: are a, b, and c consecutive integers? [#permalink]

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22 May 2010, 20:43
I got D too, used sample numbers to check. The methods given here are useful as well!!!
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Re: If a, b, and c are positive integers, with a < b < c, [#permalink]

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03 Sep 2013, 00:11
I solved the questions by plugging in consecutive numbers.
Got the ans D. Would you suggest to use plugging in method to solve such questions?

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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, 07:05
1
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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 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

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, 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, 07:41
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

Please let me know whether this makes sense.
If it does, throw me some Kudos!

Cheers
J

Notice that OA is D.
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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, 08:06
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

Please let me know whether this makes sense.
If it does, throw me some Kudos!

Cheers
J

Notice that OA is D.

Oops, typo. Yes, answer is (D) indeed

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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

1/2-1/3 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 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.

+1 D.

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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,

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.

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,

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.

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,

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.

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 (x-1), 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
gaurav1418z wrote:
Bunuel,

We can also do it by taking a, b and c as (x-1), x, (x+1) right? It looks the same to me as taking b = a+1 and c = a+2

Yes, you could do this way too.
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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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Re: If a, b, and c are positive integers, with a < b < c, [#permalink]

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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+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.

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If a, b, and c are positive integers, with a < b < c,   [#permalink] 02 Feb 2016, 19:08
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