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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] New post 26 Aug 2012, 02:12
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Re: If n is an integer, is n even? [#permalink] New post 26 Aug 2012, 02:13
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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] 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] New post 30 Aug 2012, 02:50
answer D. both statements are individually sufficient
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Re: If n is an integer, is n even? [#permalink] 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] 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] New post 31 Aug 2012, 01:33
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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] 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] New post 16 Mar 2014, 22:45
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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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COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: If n is an integer, is n even? [#permalink] 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 + 4is 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).
Re: If n is an integer, is n even?   [#permalink] 18 Mar 2014, 05:45
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