If i and j are integers, is i + j an even integer?

SOLUTION

If i and j are integers, is i + j an even integer?

(1) i < 10. Clearly insufficient.

(2) i = j:

\(i + j=i+i=2i=2*integer=even\). Sufficient.

Answer: B.
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In Easiest form
1) i<10 7 + 12 & 7+ 13 even & odd
2) i = j both either 2k or 2k+1 .. 2k + 2k = eveen ..... 2k+1 + 2k+1 = 4k+2 = even.

What if both i and j are negative ? For example -3+(-3)= -6. Can we categorise negative numbers into even or odd ? Shouldn't the answer be E?

Yes.

1. An even number is an integer that is "evenly divisible" by 2, i.e., divisible by 2 without a remainder. Since -6 IS divisible by 2 (-6/2 = -3), then -6 is an even integer. Even integers are: ..., -6, -4, -2, 0, 2, 4, 6, 8, ...

2. An odd number is an integer that is not evenly divisible by 2.


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We need to determine whether i + j is an even integer. Remember:

even + even = even

odd + odd = even

Thus, if we can determine that i and j are either both even or both odd, we will be able to answer the question.

Statement One Alone:

i < 10

Knowing that i is less than 10 is not enough to determine whether i + j is an even integer. Statement one alone is not sufficient to answer the question. We can eliminate answer choices A and D.

Statement Two Alone:

i = j

Since we know i = j, we know that i and j are both even or both odd. Following our addition rules for even and odd numbers, we see that i + j must be even.

Statement two alone is sufficient to answer the question.

The answer is B.
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I disagree... nowhere in the prompt does it specify that i and j are not equal to zero.

What am I missing?

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 (in fact zero is divisible by every integer except zero itself).

Or in another way: an even number is an integer of the form \(n=2k\), where \(k\) is an integer. So for \(k=0\) --> \(n=2*0=0\).

Also note that if we were not told that \(i\) and \(j\) are integers then this statement would not be sufficient as in this case \(j\) could be for example 1.5, so \(i+j=j+j=2j=3=odd\) or \(j\) could be for example 1.1, so \(i+j=j+j=2j=2.2\neq{integer}\).

For more on Number Properties check: https://gmatclub.com/forum/math-number-theory-88376.html

Hope it helps.
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Hi All,

We're told that I and J are integers. We're asked if(I + J) is an even integer. This is a YES/NO question. We can solve it by TESTing VALUES.

1) I < 10

IF... I = 9 and J = 1, then the answer to the question is YES.
I = 9 and J = 2, then the answer to the question is NO.
Fact 1 is INSUFFICIENT

2) I = J

Fact 2 tells us that I and J are the SAME integer, so (I+J) can be rewritten as either 2I or 2J. Since the product of any integer and 2 results in an EVEN integer, the answer to the question is ALWAYS YES. You can also prove it with examples:
I = 1 and J = 1 ... total = 2
I = 2 and J = 2 ... total = 4
I = 3 and J = 3 ... total = 6
Etc.
Fact 2 is SUFFICIENT

Final Answer:

GMAT assassins aren't born, they're made,
Rich

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

Since we have 2 variables (i and j) and 0 equations, C is most likely to be the answer and so we should consider 1) & 2) first.

Conditions 1) & 2)

i + j = i + i = 2i.
It is an even integer.

Since this question is a statistics question (one of the key question areas), CMT (Common Mistake Type) 4(A) of the VA (Variable Approach) method tells us that we should also check answers A and B.

Condition 1)
i = 1, j = 9 : i + j = 10 is an even integer.
i = 1, j = 8 : i + j = 9 is not an even integer.
The condition 1) is not sufficient.

Condition 2)
i + j = i + i = 2i.
It is an even integer.
The condition 2) only is sufficient.

Therefore, the answer is B.

Normally, in problems which require 2 or more additional equations, such as those in which the original conditions include 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, each of conditions 1) and 2) provide an additional equation. In these problems, the two key possibilities are that C is the answer (with probability 70%), and E is the answer (with probability 25%). Thus, there is only a 5% chance that A, B or D is the answer. This occurs in common mistake types 3 and 4. Since C (both conditions together are sufficient) is the most likely answer, we save time by first checking whether conditions 1) and 2) are sufficient, when taken together. Obviously, there may be cases in which the answer is A, B, D or E, but if conditions 1) and 2) are NOT sufficient when taken together, the answer must be E.

Statement 1:

Clearly Insufficient.

Statement 2:
i = j

2+2=4
3+3=6
1001+1001=2002

Sum is always even.

Sufficient.

Answer B

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