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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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If n is an integer, is n even?

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

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

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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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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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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? [#permalink]

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New post 27 Aug 2012, 08:07
If n is an integer, is n even?

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



If n^2 - 1 is odd ,then n^2 is even.Assuming both Positive and Negative integers fall under even category,I ca say n is even.

So statement 1 is alone sufficient.


If 3n + 4 is an even integer,then 3n is also even. Since 3 is odd, n has to be even.

So statement 2 is also sufficient.

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

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New post 30 Aug 2012, 10:32
(1) n^2 - 1 is odd. The next consecutive integer is n^2 and is therefore even. This means that n must be even too, because squaring a number does NOT change this. --> sufficient

(2) 3n + 4 is even. So 3n is even, too. This means that the prime factorization of 3n includes at least one 2. Dividing by 3 (to get from 3n to n) does NOT eliminate recude the number of twos in the prime factorization, so n is even. --> sufficient

The correct answer is D. Both statements are individually sufficient.

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

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New post 30 Aug 2012, 18:56
i will go with d
both options individually can get the required info
-as its an integer no more fractions
integer when squared gives the same type value i.e odd gives odd and even gives even
same way multiplying odd with even gives even and odd with odd gives odd

hence both are individually sufficient
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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 18 Mar 2014, 05:45
If n is an integer, is n even?

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


Given that, n is an integer.

Statement (1)

\(n^2 - 1=(n-1)(n+1)\) is odd =>\((n-1)\) and \((n+1)\) both are odd => \(n\) is even......Sufficient ....A B C D E

Statement (2)

\(3n + 4\) is even =>\(3n + 3\) is odd => \(3(n+1)\) is odd =>\(n + 1\) is odd =>\(n\) is even ........Sufficient....A B C D E

Since, each statement alone is sufficient, answer is (D).

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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

Option D

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

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

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New post 22 Aug 2016, 01:18
We need to find if x is even or odd
given => x is an integer
statement 1 =>n^2-1=odd => n^2=> even => n must be even
Rule used => POWER Does not effect the even / odd nature of any number

statement 2 => 3n+4=even=> 3n=even => n must even
Rulw => if XY=even => at-least one of them must be even
Hence suff

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

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New post 03 Dec 2016, 17:23
Here is my solution ->
Her we need to get the Even/Odd nature of integer n
Lets dive into statements
Statement 1
n^2-1=odd
so n^2 must be even
RULE-> POWER Does not affect the even/odd nature of any integer
Hence n must be even too.
Hence Sufficient
Statement 2->
3n+4=even
Hence 3n=even-even=even
As 3n=even and 3 is odd => n must be even to make 3n even
Hence Sufficient
Hence D

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

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New post 01 Apr 2017, 09:19
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.

Practice Questions
Question: 27
Page: 277
Difficulty: 600


(1) The even odd nature of n=n^2 (e.g. 2=Even and 2^2=4=Even)
Therefore just writing n-1= Odd
Even- Odd=Odd
1 is odd. Therefore n is even.

Sufficient. (AD)

(2) 3n+4=Even
Only two possible cases:
Even+Even= Even
Odd+Odd=Even

4 is even and the even or odd nature of 3n depends on n.
Thus n is even. (Therefore 3n=Even)

Sufficient.

D

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Re: If n is an integer, is n even?   [#permalink] 01 Apr 2017, 09:19
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