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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. 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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