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Re: If n is an integer, is n even? [#permalink]
26 Aug 2012, 02:13

Expert's post

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.

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

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

Re: If n is an integer, is n even? [#permalink]
31 Aug 2012, 01:33

Expert's post

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.

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

Re: If n is an integer, is n even? [#permalink]
16 Mar 2014, 22:45

Expert's post

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. _________________