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

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

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New post 18 Nov 2014, 08:43
1
3
00:00
A
B
C
D
E

Difficulty:

  65% (hard)

Question Stats:

66% (02:34) correct 34% (02:40) wrong based on 174 sessions

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

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New post 18 Nov 2014, 08:53
1
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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Re: If a, b, and c are positive integers, with a < b < c, are a, b, and c  [#permalink]

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

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

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New post 18 Nov 2014, 23:03
1
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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Re: If a, b, and c are positive integers, with a < b < c, are a, b, and c  [#permalink]

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New post 19 Nov 2014, 07:44
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Bunuel wrote:

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.


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

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New post 26 Feb 2018, 02:22
1/n -1/(n+1) = 1/n(n+1) --> never happens (1), (2) and (1+2) => D
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Re: If a, b, and c are positive integers, with a < b < c, are a, b, and c   [#permalink] 26 Feb 2018, 02:22
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