If n is an integer, is n + 1 odd?

SOLUTION

If n is an integer, is n + 1 odd?

In order \(n+1\) to be odd, \(n\) must be even. So, the question basically asks whether \(n\) is an even number.

(1) n + 2 is an even integer --> \(n+2=even\) --> \(n=even-2=even\). Sufficient.
(2) n -1 is an odd integer --> \(n-1=odd\) --> \(n=odd+1=even\). Sufficient.

Answer: D.
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A. N, N+1, N+2. If N+2 is even, then N+1 is odd
Sufficient

B. N-1, N, N+1. If N-1 is odd, then N+1 is odd.
Sufficient

The following Rules will be handy in order to solve this question:

Even Number (+/-) Odd Number = Odd Number------->(a)
Even Number (+/-) Even Number = Even Number------>(b)
Odd Number (+/-) Odd Number = Even Number.------->(c)

Now coming to the question stem: is N + 1 = Odd. From the question stem itself we know that 1 is an odd number and that the result is odd. From equation (a) above we know that this is only possible when n is even. So, the question is actually asking as if n is even?

Statement (1): n + 2`= Even. "2" as we know is even so this result is only possible if n is even (b). SUFFICIENT

Statement (2): n- 1 = Odd. "1" is an Odd Number. This result is only possible if n is even (b). SUFFICIENT

Ans (D)

SOLUTION

If n is an integer, is n + 1 odd?

In order \(n+1\) to be odd, \(n\) must be even. So, the question basically asks whether \(n\) is an even number.

(1) n + 2 is an even integer --> \(n+2=even\) --> \(n=even-2=even\). Sufficient.
(2) n -1 is an odd integer --> \(n-1=odd\) --> \(n=odd+1=even\). Sufficient.

Answer: D.
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statement 1 :n+2=even, n=even, n+1 == odd, sufficient
Statement 2; n-1=odd,n=even, n+1 == odd, sufficient

hence D

We need to determine whether n + 1 is odd. Remember, if n is even, then n + 1 will be odd, and if n is odd, n + 1 will be even.

Statement One Alone:

n + 2 is an even integer.

Statement one tells us that n is an even integer (since 2 more than an even integer is also an even integer). Because n is an even integer, we know that n + 1 is odd. Statement one is sufficient to answer the question. We can eliminate answer choices B, C, and E.

Statement Two Alone:

n – 1 is an odd integer.

Since subtraction rules for even and odd integers are the same as addition rules for even and odd integers, we know that if n – 1 is odd, then n itself must be even, and hence n + 1 must also be odd. Statement two is sufficient to answer the question.

Answer: D
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Here is my solution->
Some Basic Rules before we attempt this Even/odd Question

Even+Even=Even
Even+Odd=Odd
Odd+Even=Odd
Odd+Odd=Even

Similarly

Even-Even=Even
Even-Odd=Odd
Odd-Even=Odd
Odd-Odd=Even


The Question is asking us whether n+1 is odd or not
n+1 will be odd only when n is even.
Hence question can be rephrased into "is n even"
Statement 1
n+2 is even => n=even -2 => even
Hence sufficient
Statement 2
n-1=odd
Hence n=odd+1=> even

hence sufficient
Hence D
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Target question: Is n + 1 odd?

Some important rules:
#1. ODD +/- ODD = EVEN
#2. ODD +/- EVEN = ODD
#3. EVEN +/- EVEN = EVEN


Statement 1: n + 2 is an even integer.
In other words: n + EVEN = EVEN
In other words: n = EVEN - EVEN
By Rule #3, n must be even, which means n+1 is ODD
The answer to the target question is YES, n+1 IS odd
Since we can answer the target question with certainty, statement 1 is SUFFICIENT

Statement 2: n -1 is an odd integer.
In other words: n - ODD = ODD
In other words: n = ODD + ODD
By Rule #1, n must be even, which means n+1 is ODD
The answer to the target question is YES, n+1 IS odd
Since we can answer the target question with certainty, statement 2 is SUFFICIENT

Answer: D

Cheers,

#1
n+2 even ; n has to be even so n+1 is odd sufficient
#2
n-1 is odd integer ; n = even integer so
n+1 is odd; sufficient
IMO D

Forget the conventional way to solve DS questions.

We will solve this DS question using the variable approach.

Remember the relation between the Variable Approach, and Common Mistake Types 3 and 4 (A and B)[Watch lessons on our website to master these approaches and tips]

Step 1: Apply Variable Approach(VA)

Step II: After applying VA, if A or B is the answer, check whether the question is key questions.

StepIII: If the question is not a key question, choose A or B as the probable answer, but if the question is a key question, apply CMT 3 and 4 (A or B).

Step IV: If CMT3 or 4 (A or B) is applied, choose either A, B, or D.

Let's apply CMT (2), which says there should be only one answer for the condition to be sufficient. Also, this is an integer question and, therefore, we will have to apply CMT 3 and 4 (A or B).

To master the Variable Approach, visit https://www.mathrevolution.com and check our lessons and proven techniques to score high in DS questions.

Let’s apply the 3 steps suggested previously. [Watch lessons on our website to master these 3 steps]

Step 1 of the Variable Approach: Modifying and rechecking the original condition and the question.

We have to find whether 'n + 1' is odd.

=> n + 1 is odd therefore, 'n has to be even. So, we need to find whether 'n' is even- where 'n' is an integer.

Second and the third step of Variable Approach: From the original condition, we have 1 variable (n and n). To match the number of variables with the number of equations, we need 1 equation. Since conditions (1) and (2) will provide 1 equation each, D would most likely be the answer.

But we know that this is a key question [Integer question] and if we get an easy A or B as an answer, we will choose D.

Let’s take a look at each condition.

Condition(1) tells us that n + 2 is an even integer .

=> for n + 2 to be been , 'n' has to be even - YES


Since the answer is unique YES, the condition is sufficient by CMT 1.


Condition(2) tells us that 'n -1' is an odd integer.

=> n - 1 = odd => n = odd + 1 = even - YES


Since the answer is a unique YES , the condition is sufficient by CMT 1.


Both conditions alone are sufficient.

So, D is the correct answer.

Answer: D

Given is the fact that n is an integer, meaning it can be negative or positive. In statement 1, it says n+2 = even, n+2 can also be 0 as zero is even, this leaves as n = -2, now in this case n+1 = -2+1 = -1, which is not an odd number(odd numbers are whole numbers not divisible by 2 starting 1,3,5 ....). So we get an answer NO for this example. Another case could be n+2 = 8, meaning n = 6, n+1 = 6+1 = 7 which is odd, YES. Hence A is not sufficient. Why 0 case is not considered?

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