Wicked question by MGMAT this week! Check it out...
MGMAT Challenge Problem of the Week
10/19/09
Question
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
(A)
 Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.

(B)
 Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.

(C)
 BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.

(D)
 EACH statement ALONE is sufficient.

(E)
 Statements (1) and (2) TOGETHER are NOT sufficient to answer the question asked, and additional data are needed.

Answer
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
1/
a – 1/(
a + 1) = 1/
c
(
a + 1)/[
a(
a + 1)] –
a/[
a(
a + 1)] = 1/
c
1//[
a(
a + 1)] = 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 (1/
a – 1/
b = 1/
c). The first 4 possibilities are as follows:
1/1 – 1/2 = 1/2
1/2 – 1/3 = 1/6
1/3 – 1/4 = 1/12
1/4 – 1/5 = 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
2
a + 2 =
a^{2} + 2
a
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.
The answer is D: Each statement is sufficient.

The first option required just a little more calculation that the second.
I did get the answer on my first try, but I still wanted to post it here because the logic that struck me while trying to solve it may not have struck me at another point of time. I know I've opened my mind to newer ways of solving questions. Earlier, even if the idea would've struck me, just knowing it was a little bit more complicated than what I'm extremely comfortable with, I would've confused myself. Now, I take a deep breath, try to understand why I'm thinking that way, and the answer comes. Even if it doesn't, atleast I'm aware of the logic I used! I should be thanking the people at T.I.M.E classes for this. The Math teacher in Delhi (Kailash Colony) as well as Bombay (Charni Road) were pretty damn good. And they have helped me think of Math in a simpler, clearer manner, and I think I'm getting over my fear now, and beginning to not underestimate myself.
I think I've left my notepad at home. I should
write my thought process here.