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# S98-15

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Math Expert
Joined: 02 Sep 2009
Posts: 53005

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16 Sep 2014, 00:52
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Difficulty:

65% (hard)

Question Stats:

54% (02:24) correct 46% (02:09) wrong based on 24 sessions

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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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Joined: 02 Sep 2009
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16 Sep 2014, 00:52
Official Solution:

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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06 Feb 2019, 01:54
Why can't I just put the consecutive values like a=1, b=2 and c=3
and then try to find out if LHS=RHS
But that doesn't work, please explain where m i missing?
Re: S98-15   [#permalink] 06 Feb 2019, 01:54
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# S98-15

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