|GRE Number Theory
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|Author:||questioner [ Tue Dec 18, 2012 7:49 am ]|
|Post subject:||GRE Number Theory|
a and b are consecutive positive integers and b > a. Which of the following statements will always be true?
Indicate ALL such statements.
(a + b) is odd
(b – a) is odd
(b² – a²) is even
(b² + 1 + a²) is even
(b – a)³ is odd
The correct answers are (a + b) is odd, (b – a) is odd, (b² + 1 + a²) is even, and (b – a)³ is odd.
Let a be represented by n; let b be represented by (n+1), since a and b are consecutive integers.
Because they are consecutive, we also know that one of them is even and the other one is odd. It is also important to note that any time we multiply any integer by 2, the resulting value will be an even number.
Let's now evaluate each answer choice.
The first answer choice:
a + b = n + (n + 1) = 2n + 1. Since 2n is even, then 2n + 1 is odd.
The second answer choice:
(b – a) = (n + 1) – n = 1, which is odd.
The third answer choice:
b² – a² = (n+1)² – n² = n² + 2n + 1 – n² = 2n + 1 = odd (see earlier explanation).
The fourth answer choice:
(b² + 1 + a²) = (n + 1)² + 1 + n² = n² + 2n + 1 + 1 + n² = 2n² + 2n + 2 is an even number.
The fifth answer choice:
(b – a)³ = (n + 1 – n)³ = 1³ = 1 is odd.
The answer you provided is wrong. (b² – a²) is odd. Please check it again. I am quite certain about it.
|Author:||Gennadiy [ Tue Dec 18, 2012 7:56 am ]|
|Post subject:||Re: GRE Number Theory|
The answer you provided is wrong. (b² – a²) is odd. Please check it again. I am quite certain about it.(b² – a²) is odd indeed. That is why the statement "(b² – a²) is even" is false and is NOT one of the credited answer choices.
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