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If n is an integer, is n even? [#permalink]
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26 Aug 2012, 03:12
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Re: If n is an integer, is n even? [#permalink]
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26 Aug 2012, 03:13
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SOLUTION:If n is an integer, is n even?(1) n^2  1 is an odd integer > \(n^21=odd\) > \(n^2=odd+1=even\). 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 > \(3n + 4=even\) > \(3n=even4=even\). 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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31 Aug 2012, 02:33
SOLUTION:If n is an integer, is n even?(1) n^2  1 is an odd integer > \(n^21=odd\) > \(n^2=odd+1=even\). 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 > \(3n + 4=even\) > \(3n=even4=even\). 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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27 Aug 2012, 09:07
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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30 Aug 2012, 03:50
answer D. both statements are individually sufficient
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Re: If n is an integer, is n even? [#permalink]
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30 Aug 2012, 11:32
(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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30 Aug 2012, 19:56
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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16 Mar 2014, 18:55
Bunuel wrote: SOLUTION:
If n is an integer, is n even?
(1) n^2  1 is an odd integer > \(n^21=odd\) > \(n^2=odd+1=even\). 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 > \(3n + 4=even\) > \(3n=even4=even\). 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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16 Mar 2014, 23:45
X017in wrote: Bunuel wrote: SOLUTION:
If n is an integer, is n even?
(1) n^2  1 is an odd integer > \(n^21=odd\) > \(n^2=odd+1=even\). 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 > \(3n + 4=even\) > \(3n=even4=even\). 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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18 Mar 2014, 06:45
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)
\(n^2  1=(n1)(n+1)\) is odd =>\((n1)\) 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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09 Dec 2015, 00:06
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 = (n1)(n+1) = odd. This means both n1 and n+1 are odd Odd*Odd = Odd Odd*Even = Even Even*Even = Evenn1, n, n+1 are three consecutive integers. Since we know that both n1 and n+1 are odd Hence n has to be even.SUFFICIENTStatement 2: 3n + 4 is an even integer Even + Even = Even Even + Odd = Odd Odd + Odd + OddSince 3n+4 = even and 4 is an even integer. Hence 3n = even. Therefore n = even SUFFICIENTOption D
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Re: If n is an integer, is n even? [#permalink]
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09 Aug 2016, 13:33
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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If n is an integer, is n even? [#permalink]
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22 Aug 2016, 02:18
We need to find if x is even or odd given => x is an integer statement 1 =>n^21=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 => atleast 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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03 Dec 2016, 18:23
Here is my solution > Her we need to get the Even/Odd nature of integer n Lets dive into statements Statement 1 n^21=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=eveneven=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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01 Apr 2017, 10:19
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 n1= 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




Re: If n is an integer, is n even?
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