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If n is an integer, is n even?

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If n is an integer, is n even?  [#permalink]

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New post 26 Aug 2012, 02:12
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Re: If n is an integer, is n even?  [#permalink]

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New post 26 Aug 2012, 02:13
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SOLUTION:

If n is an integer, is n even?

(1) n^2 - 1 is an odd integer --> \(n^2-1=odd\) --> \(n^2=odd+1=even\). Now, since \(n\) is an integer, then in order \(n^2\) to be even \(n\) must be even. Sufficient.
Notice that if we were not told that \(n\) is an integer, then \(n\) could be some irrational number (square root of an even number), for example \(\sqrt{2}\), so not an even integer.

(2) 3n + 4 is an even integer --> \(3n + 4=even\) --> \(3n=even-4=even\). The same here, since \(n\) is an integer, then in order \(3n\) to be even \(n\) must be even. Sufficient.
Notice that if we were not told that \(n\) is an integer, then \(n\) could be some fraction, for example \(\frac{2}{3}\), so not an even integer.

Answer: D.
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Re: If n is an integer, is n even?  [#permalink]

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New post 16 Mar 2014, 17:55
Bunuel wrote:
SOLUTION:

If n is an integer, is n even?

(1) n^2 - 1 is an odd integer --> \(n^2-1=odd\) --> \(n^2=odd+1=even\). Now, since \(n\) is an integer, then in order \(n^2\) to be even \(n\) must be even. Sufficient.
Notice that if we were not told that \(n\) is an integer, then \(n\) could be some irrational number (square root of an even number), for example \(\sqrt{2}\), so not an even integer.

(2) 3n + 4 is an even integer --> \(3n + 4=even\) --> \(3n=even-4=even\). The same here, since \(n\) is an integer, then in order \(3n\) to be even \(n\) must be even. Sufficient.
Notice that if we were not told that \(n\) is an integer, then \(n\) could be some fraction, for example \(\frac{2}{3}\), so not an even integer.

Answer: D.


Since n is a integer, can we not try with n as 0?
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Re: If n is an integer, is n even?  [#permalink]

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New post 16 Mar 2014, 22:45
X017in wrote:
Bunuel wrote:
SOLUTION:

If n is an integer, is n even?

(1) n^2 - 1 is an odd integer --> \(n^2-1=odd\) --> \(n^2=odd+1=even\). Now, since \(n\) is an integer, then in order \(n^2\) to be even \(n\) must be even. Sufficient.
Notice that if we were not told that \(n\) is an integer, then \(n\) could be some irrational number (square root of an even number), for example \(\sqrt{2}\), so not an even integer.

(2) 3n + 4 is an even integer --> \(3n + 4=even\) --> \(3n=even-4=even\). The same here, since \(n\) is an integer, then in order \(3n\) to be even \(n\) must be even. Sufficient.
Notice that if we were not told that \(n\) is an integer, then \(n\) could be some fraction, for example \(\frac{2}{3}\), so not an even integer.

Answer: D.


Since n is a integer, can we not try with n as 0?


Yes, n can be 0 but 0 is even too.
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Re: If n is an integer, is n even?  [#permalink]

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New post 08 Dec 2015, 23:06
Bunuel wrote:
If n is an integer, is n even?

(1) n^2 - 1 is an odd integer.
(2) 3n + 4 is an even integer.



Given: n is an integer
Required: is n even?

Statement 1: \(n^2\) - 1 is an odd integer
\(n^2\) - 1 = (n-1)(n+1) = odd.
This means both n-1 and n+1 are odd
Odd*Odd = Odd
Odd*Even = Even
Even*Even = Even


n-1, n, n+1 are three consecutive integers.
Since we know that both n-1 and n+1 are odd
Hence n has to be even.

SUFFICIENT

Statement 2: 3n + 4 is an even integer
Even + Even = Even
Even + Odd = Odd
Odd + Odd + Odd


Since 3n+4 = even and 4 is an even integer.
Hence 3n = even. Therefore n = even
SUFFICIENT

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Re: If n is an integer, is n even?  [#permalink]

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New post 09 Aug 2016, 12:33
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Quote:
If n is an integer, is n even?

(1) n^2 - 1 is an odd integer.
(2) 3n + 4 is an even integer.


We need to determine whether integer n is even. Let's review four facts about even and odd integers: 1) An integer and its square are either both even or both odd. 2) The sum (or difference) between an even integer and an odd integer is always odd. 3) The sum of two even integers (or two odd integers) is always even. 4) If the product of two integers is even, at least one of them must be even.

Statement One Alone:

(n^2) - 1 is an odd integer.

Since (n^2) - 1 is an odd integer, we know that n^2 must be even and thus n must be even. Statement one is sufficient to answer the question. We can eliminate answer choices B, C, and E.

Statement Two Alone:

3n + 4 is an even integer.

Since 3n + even = even integer, we know that 3n must be even, and since 3 is odd, n must be even. Statement two is sufficient to answer the question.

The answer is D.
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Re: If n is an integer, is n even?  [#permalink]

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New post 09 Dec 2017, 06:33
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Bunuel wrote:
If n is an integer, is n even?

(1) n^2 - 1 is an odd integer.
(2) 3n + 4 is an even integer.


Some important rules:
1. ODD +/- ODD = EVEN
2. EVEN +/- ODD = ODD
3. EVEN +/- EVEN = EVEN

4. (ODD)(ODD) = ODD
5. (ODD)(EVEN) = EVEN
6. (EVEN)(EVEN) = EVEN



Target question: Is integer n EVEN?

Statement 1: n² - 1 is an odd integer
n² - 1 = (n + 1)(n - 1)
So, statement 1 is telling us that (n + 1)(n - 1) = ODD
From rule #4 (above), we can conclude that BOTH (n + 1) and (n - 1) are ODD
If (n + 1) is ODD, then n must be EVEN (since 1 is ODD, we can apply rule #2 to conclude that n is EVEN)
If (n - 1) is ODD, then n must be EVEN (by rule #2 )
So, the answer to the target question is YES, n is even
Since we can answer the target question with certainty, statement 1 is SUFFICIENT

Statement 2: 3n + 4 is an even integer
In other words, (3n + EVEN) is EVEN
From rule #3, we can conclude that 3n is EVEN

Since 3 is odd, we can write: (ODD)(n) = EVEN
From rule #5, we can conclude that n is EVEN
So, the answer to the target question is YES, n is even
Since we can answer the target question with certainty, statement 2 is SUFFICIENT

Answer: D

Cheers,
Brent
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Re: If n is an integer, is n even? &nbs [#permalink] 09 Dec 2017, 06:33
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