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If n is an integer, is n even?
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26 Aug 2012, 03:12
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85% (01:06) correct 15% (01:17) wrong based on 1820 sessions
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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. Practice Questions Question: 27 Page: 277 Difficulty: 600
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Re: If n is an integer, is n even?
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26 Aug 2012, 03:13
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?
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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?
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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?
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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?
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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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Re: If n is an integer, is n even?
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09 Dec 2017, 07:33
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) = EVENTarget 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 evenSince we can answer the target question with certainty, statement 1 is SUFFICIENT Statement 2: 3n + 4 is an even integerIn 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 EVENSo, the answer to the target question is YES, n is evenSince 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?
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