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SMAbbas
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If a, b, and c are positive integers, with a < b < c, are a, b, and c Updated on: 02 Oct 2020, 02:07
Originally posted by SMAbbas on 22 Oct 2009, 23:27.
Last edited by Bunuel on 02 Oct 2020, 02:07, edited 2 times in total.
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If \(a\), \(b\), and \(c\) are positive integers, with \(a \lt b \lt c\), are \(a\), \(b\), and \(c\) consecutive integers?


(1) \(\frac{1}{a} - \frac{1}{b} = \frac{1}{c}\)

(2) \(a + c = b^2 - 1\)
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D
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If a, b, and c are positive integers, with a < b < c, are a, b, and c 22 Oct 2009, 23:59
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SMAbbas
If \(a\), \(b\), and \(c\) are positive integers, with \(a \lt b \lt c\), are \(a\), \(b\), and \(c\) consecutive integers?


(1) \(\frac{1}{a} - \frac{1}{b} = \frac{1}{c}\)

(2) \(a + c = b^2 - 1\)
Show more


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.
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If a, b, and c are positive integers, with a < b < c, are a, b, and c 18 Nov 2014, 09:24
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Hi Bunuel,

Answer is D. For the consecutive numbers; b=(a+c)/2.

Statement A, a+c= b^2 -1. But for consecutive, it should be a+c= 2b. Hence not consecutive.
From B, b=ca/(c-a). Hence, the numbers are not consecutive.
Both stat suff. Answer D.
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If a, b, and c are positive integers, with a < b < c, are a, b, and c 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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If a, b, and c are positive integers, with a < b < c, are a, b, and c Updated on: 24 Oct 2013, 08:05
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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SMAbbas
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
Show more

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 :)
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If a, b, and c are positive integers, with a < b < c, are a, b, and c 14 May 2014, 04:40
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Bunuel
SMAbbas
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 :?: :shock:

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.
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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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If a, b, and c are positive integers, with a < b < c, are a, b, and c 14 May 2014, 06:35
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Bunuel
SMAbbas
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 :?: :shock:

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 ?
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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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If a, b, and c are positive integers, with a < b < c, are a, b, and c 18 Nov 2014, 08:53
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Not sure if this was the correct approach or not...

statement 1: sufficient
If we say a=1 b=2 c=3 then this equation would be 1/1 - 1/2 = 1/3 which is not true. Also seems to be right for all the other quick consecutive #s I plugged in. I'm going to say that this is sufficient by showing us that they are not consecutive.

statement 2: sufficient
If we say a=1 b=2 c=3 then this equation would be 1+3=2^2 - 1
4=4-1
4=3
That also does not work so I'm going to say that this is sufficient for the same reason as above.

Answer D!
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If a, b, and c are positive integers, with a < b < c, are a, b, and c 18 Nov 2014, 13:24
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I tried the problem in the following way:

let a = n-1, b = n and c = n+1, assuming that the numbers are consecutive.

from statement 1-
1/(n-1) - 1/n = 1/(n+1); this reduces to n^2-2n-1 = 0. This equation does not have any integer roots. i.e. n = 1+sqrt(2) or n = 1- sqrt(2). so consecutive numbers do not satisfy this equation. sufficient to prove that a,b and c are not consecutive.

from statement -2:
n + 1 = n^2 - n, this also reduces to n^2-2n-1 = 0. same reasoning as above. sufficient.

therefore answer is : [D]
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If a, b, and c are positive integers, with a < b < c, are a, b, and c 18 Nov 2014, 23:03
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testing both the statements with simple consecutive no.1,2 & 3

statement 1: 1/a -1/b =1/c
Clearly it shows a,b &c are not consecutive.


statement 2: a+c=b^2-1
This statement too gives the definite result that the no.s are not consecutive.


Answer is D
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If a, b, and c are positive integers, with a < b < c, are a, b, and c 19 Nov 2014, 07:44
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Bunuel

Tough and Tricky questions: Algebra.



If \(a\), \(b\), and \(c\) are positive integers, with \(a \lt b \lt c\), are \(a\), \(b\), and \(c\) consecutive integers?


(1) \(\frac{1}{a} - \frac{1}{b} = \frac{1}{c}\)

(2) \(a + c = b^2 - 1\)

Kudos for a correct solution.
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Official Solution:

If \(a\), \(b\), and \(c\) are positive integers, with \(a \lt b \lt c\), are \(a\), \(b\), and \(c\) consecutive integers?

The question can be rephrased "Is \(b = a + 1\) and is \(c = a + 2\)?"

One way to approach the statements is to substitute these expressions involving \(a\) and solve for \(a\). Since this could involve a lot of algebra at the start, we can just substitute \(a + 1\) for \(b\) and test whether \(c = a + 2\), given that both are integers.

Statement 1: SUFFICIENT.

Following the latter method, we have
\(\frac{1}{a} - \frac{1}{a + 1} = \frac{1}{c}\)
\(\frac{a + 1}{a(a + 1)} - \frac{a}{a(a + 1)} = \frac{1}{c}\)
\(\frac{1}{a(a + 1)} = \frac{1}{c}\)
\(a^2 + a = c\)

Now we substitute a + 2 for c and examine the results:
\(a^2 + a = a + 2\)
\(a^2 = 2\)

\(a\) is the square root of 2. However, since \(a\) is supposed to be an integer, we know that our assumptions were false, and \(a\), \(b\), and \(c\) cannot be consecutive integers.

We can now answer the question with a definitive "No," making this statement sufficient.

We could also test numbers. Making \(a\) and \(b\) consecutive positive integers, we can solve the original equation \((\frac{1}{a} - \frac{1}{b} = \frac{1}{c})\). The first 4 possibilities are as follows:
\(\frac{1}{1} - \frac{1}{2} = \frac{1}{2}\)
\(\frac{1}{2} - \frac{1}{3} = \frac{1}{6}\)
\(\frac{1}{3} - \frac{1}{4} = \frac{1}{12}\)
\(\frac{1}{4} - \frac{1}{5} = \frac{1}{20}\)

Examining the denominators, we can see that \(c = ab\). None of these triples so far are consecutive, and as \(a\) and \(b\) get larger, \(c\) will become more and more distant, leading us to conclude that \(a\), \(b\), and \(c\) are not consecutive.

Statement 2: SUFFICIENT.

Let's try substituting \((a + 1)\) for \(b\) and \((a + 2)\) for \(c\).
\(a + a + 2 = (a + 1)^2 - 1\)
\(2a + 2 = a^2 + 2a\)
\(2 = a^2\)

Again, we get that \(a\) must be the square root of 2. However, we know that \(a\) is an integer, so the assumptions must be false. We can answer the question with a definitive "No," and so the statement is sufficient.

Answer: D.
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If a, b, and c are positive integers, with a < b < c, are a, b, and c 07 Jun 2020, 02:05
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The question says that a, b & c are positive integers.
To verify if they are consecutive, we can substitute 'a+1' in place of b and 'a+2' in place of c in each of the equations.

(1) Solution a=sq rt. of 2. As the value of a as per (1) is not an integer, a, b and c cannot be consecutive integers.
(2) Solution a=sq rt. of 2. As the value of a as per (2) is not an integer, a, b and c cannot be consecutive integers.
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If a, b, and c are positive integers, with a < b < c, are a, b, and c 10 Feb 2022, 09:50
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IanStewart check out this cool question :)
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If a, b, and c are positive integers, with a < b < c, are a, b, and c 19 Jul 2025, 18:04
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