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(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.

(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.

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 _________________

(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.

(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.

(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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(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]
09 Dec 2015, 00:06

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