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If n is an integer, is n even?

(1) 2n is divisible by 4
(2) n^2 is even

Hello Bunuel, I got confused when I was solving this question. First let's agree on a couple of points, 0 is an integer, 0 is neither even nor odd, & 0 is divisible by everything.

So when I look at the first statement it says 2n is divisible by 4, what if n=0, this statement would still be correct, however 0 is neither even nor odd, so how is statement one sufficient on it's own.

Regards,
YA
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Hi youssefhabouseif,

0 IS an EVEN integer, by definition. Here's why:

On a number line, consecutive integers fall into the pattern....even....odd.....even....odd.....even....odd....etc.

-2 = even
-1 = odd
0 = even
1 = odd
2 = even
Etc.

Knowing that 0 IS EVEN, you likely would have picked the correct answer, but it's important to keep this rule in mind - you'll likely use it on Test Day at least once.

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youssefhabouseif
If n is an integer, is n even?

(1) 2n is divisible by 4
(2) n^2 is even

Hello Bunuel, I got confused when I was solving this question. First let's agree on a couple of points, 0 is an integer, 0 is neither even nor odd, & 0 is divisible by everything.

So when I look at the first statement it says 2n is divisible by 4, what if n=0, this statement would still be correct, however 0 is neither even nor odd, so how is statement one sufficient on it's own.

Regards,
YA

ZERO:

1. 0 is an integer.

2. 0 is an even integer. An even number is an integer that is "evenly divisible" by 2, i.e., divisible by 2 without a remainder and as zero is evenly divisible by 2 then it must be even.

3. 0 is neither positive nor negative integer (the only one of this kind).

4. 0 is divisible by EVERY integer except 0 itself.

Check more here: tips-and-hints-for-specific-quant-topics-with-examples-172096.html#p1371030
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Here given n is an integer we need to see if its even or odd
Statement 1 => 2n/4=Integer => N is divisible by 2 => Its even [[as The numbers that are divisible by 2 are evens ]]
Statement 2 => n^2=even => n must be even as POWER DOES NOT EFFECT THE EVEN OR ODD NATURE OF ANY NUMBER.
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Bunuel
If n is an integer, is n even?

(1) 2n is divisible by 4 --> 2n = 4k --> n = 2k = even. Sufficient.

(2) n^2 is even --> since n is an integer, then n^2 can be even only if n is even. Sufficient.

Answer: D.


For the 2nd case, what if n is negative and n^2 is even. ex: n=-2. then n^2=4. should we consider n (which is negative as even).
The question boils down to whether negative numbers are considered odd/even in GMAT?
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Bunuel
If n is an integer, is n even?

(1) 2n is divisible by 4 --> 2n = 4k --> n = 2k = even. Sufficient.

(2) n^2 is even --> since n is an integer, then n^2 can be even only if n is even. Sufficient.

Answer: D.


For the 2nd case, what if n is negative and n^2 is even. ex: n=-2. then n^2=4. should we consider n (which is negative as even).
The question boils down to whether negative numbers are considered odd/even in GMAT?

Negative integers can be even or odd generally, not only on the GMAT. GMAT does not have its own math.

An even number is an integer that is "evenly divisible" by 2, i.e., divisible by 2 without a remainder. So, ..., -4, -2, 0, 2, 4, ... are all even integers.

An odd number is an integer that is not evenly divisible by 2. So, ..., -3, -1, 1, 3, 5, ... are all odd integers.
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From the question stem, we know that n is not a fraction or a root.

From statement I alone, 2n is divisible by 4.

Therefore, 2n = 4, 8, 12, …. and so on. This means n = 2 or 4 or 6 and so on.
For all values, we see that n is even
Statement I alone is sufficient. Answer options B, C and E can be eliminated.

From statement II alone, \(n^2\) is even.

The square of an integer is even only when the integer is even. Therefore, n is even.
Statement II alone is sufficient. Answer option A can be eliminated.

The correct answer option is D.

Hope that helps!
Aravind B T
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What if n is 13? A would be not sufficient?

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What if n is 13? A would be not sufficient?

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(1) says that 2n is divisible by 4. If n = 13, 2n=26, which is not divisible by 4, so n cannot be 13.
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In the statement 1, what if we assume if n=1, then also 2n is divisible by 4, hence when n is odd then also it satisfies, and if n=2 then also it satisfies. Isn't it insufficient in that case?
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In the statement 1, what if we assume if n=1, then also 2n is divisible by 4, hence when n is odd then also it satisfies, and if n=2 then also it satisfies. Isn't it insufficient in that case?

If n = 1, then 2n = 2, which is not divisible by 4. So your assumption there is incorrect. For 2n to be divisible by 4, n itself must be even, because then n contributes at least one factor of 2, making 2n divisible by 4.
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