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Math Expert V
Joined: 02 Sep 2009
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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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Difficulty:   65% (hard)

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

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

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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.

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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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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.
Senior Manager  P
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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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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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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.

Math Expert V
Joined: 02 Sep 2009
Posts: 58434
Re: If a, b, and c are positive integers, with a < b < c, are a, b, and c  [#permalink]

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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.

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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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1/n -1/(n+1) = 1/n(n+1) --> never happens (1), (2) and (1+2) => D 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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