Is the sum of four particular integers even?

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Is the sum of four particular integers even?

(1) Two of the integers are odd and two are even.
(2) The average (arithmetic mean) of the four integers is an integer.

D

Source: Official Guide 2020

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by Ian Stewart » Sun Apr 28, 2019 12:08 am

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Statement 1 tells us exactly what kind of integers we have, so using the familiar even/odd rules, we can work out if the sum will be even or odd (since odd+odd is even, and even+even is even, the sum will turn out to be even, though since it's a DS question, we don't actually care what the answer is, only that we can find an answer, so there's no reason to bother even doing that much work).

Statement 2 tells us that if we divide the sum by 4, we get an integer. So the sum is divisible by 4, and is thus certainly divisible by 2, which is another way of saying that it's even.

So the answer is D.
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by Brent@GMATPrepNow » Sun Apr 28, 2019 6:32 am

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AbeNeedsAnswers wrote:Is the sum of four particular integers even?

(1) Two of the integers are odd and two are even.
(2) The average (arithmetic mean) of the four integers is an integer.
Some important rules:
#1. ODD +/- ODD = EVEN
#2. ODD +/- EVEN = ODD
#3. EVEN +/- EVEN = EVEN

#4. (ODD)(ODD) = ODD
#5. (ODD)(EVEN) = EVEN
#6. (EVEN)(EVEN) = EVEN


Target question: Is the sum of four particular integers even?

Statement 1: Two of the integers are odd and two are even.
We have: SUM = ODD + ODD + EVEN + EVEN
= (ODD + ODD) + (EVEN + EVEN)
= EVEN + EVEN
= EVEN
The answer to the target question is YES, the sum is even
Since we can answer the target question with certainty, statement 1 is SUFFICIENT

Statement 2: The average (arithmetic mean) of the four integers is an integer.
Let the 4 numbers be a, b, c, and d
We can write: (a + b + c + d)/4 = k, where k is some integer
So, we can also write: a + b + c + d = 4k
Since 4k is EVEN, we know that a + b + c + d is EVEN
The answer to the target question is YES, the sum is even
Since we can answer the target question with certainty, statement 2 is SUFFICIENT

Answer: D

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by [email protected] » Sun Apr 28, 2019 11:41 am

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Hi All,

We're told that we have four INTEGERS. We're asked if the sum of those four integers is EVEN. This is a YES/NO question and can be solved with either Number Property rules or TESTing VALUES.

(1) Two of the integers are ODD and two are EVEN.

Fact 1 gives us enough information to deal with some specific Number Properties.
Adding two ODD numbers will give us a sum that is EVEN (for example, 3+5 = 8)
Adding two EVEN numbers will give us a sum that is EVEN (for example, 2+4 = 6)

Thus, adding two ODDs and two EVENs will give us EVEN+EVEN = EVEN. The answer to the question is ALWAYS YES.
Fact 1 is SUFFICIENT

(2) The average (arithmetic mean) of the four integers is an integer.

Fact 2 references the 'Average Formula', so we should think in those terms. If we refer to the values as A, B, C and D, we can set up an average as:

(A+B+C+D)/4 = Integer
(A+B+C+D) = (4)(Integer)

From this equation, we know that the sum of the four integers will be 4 times an integer. Multiplying ANY integer by an EVEN number will give us a product that is EVEN. For example:
4(-1) = -4
4(0) = 0
4(1) = 4
4(2) = 8
4(3) = 12
Etc.
Thus, the answer to the question is ALWAYS YES.
Fact 2 is SUFFICIENT

Final Answer: D

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by SampathKp » Wed Dec 18, 2019 4:41 am

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AbeNeedsAnswers wrote:Is the sum of four particular integers even?

(1) Two of the integers are odd and two are even.
(2) The average (arithmetic mean) of the four integers is an integer.

D

Source: Official Guide 2020
Sum of 2 even integers is always even and sum of 2 Odd integers is always Even. So from (1) sum of 4 particular integers of which 2 are odd and 2 are even is always even. Hence (1) is Sufficient

Arithemetic mean of 4 integers is Sum of the 4 integers/4 / For this to be an integer, the numerator has to be even becuase only even integers are perferectly divisible by 4. Hence (2) alone is Sufficient to answer the question.

Answer is D , i.e Each Statement Alone is sufficient to Answer the Question asked.