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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.

This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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

(1) i < 10 (2) i = j

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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Re: If i and j are integers, is i + j an even integer? [#permalink]

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03 Dec 2016, 17:16

Here is my solution -> Notice i and j are integers (this info is very important in even/odd Questions) We need to check the even/odd nature of i+j Notice i+j will be even when either both i and j are even or both i and j are odd Lets dive into statements Statement 1 i<10 no clue of the even /odd nature of i and j Additionally we must remember that negative integers can be even/odd too. Hence Insufficient Statement 2 i=j so i+j=2i,which is always even Hence sufficient

Re: If i and j are integers, is i + j an even integer? [#permalink]

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02 Apr 2017, 21:46

Bunel's explanation is very handy in case you don't recall the odd/even rules.

If you recall that Odd + Odd = Even and Even + Even = Even, statements two allows you to answer the question confidently without any calculation.
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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

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

(1) i < 10 (2) i = j

Practice Questions Question: 14 Page: 276 Difficulty: 550

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.
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