Is the positive integer x even?

Question

(1) (x - 1) is a prime number
(2) (x^2 - 1) is a prime number

KarishmaParmar · Answer

Statement 1- the exception is 2 so can't say
Statement 2 - (x-1) (x+1) is odd so, both these terms should be odd (since they are alternate consecutive, they have to be the same type odd or even )and x would be even

Answer B

OptimusPrepJanielle · Answer

Statement 1: (x - 1) is a prime number
x = 3 implies (x - 1) = 2. A prime number
x = 4 implies (x - 1) = 3. A prime number.
Hence x can be both even and odd. Insufficient

Statement 2: (x^2 - 1) is a prime number
(x - 1)(x + 1) is a prime number.
The only possible value of (x - 1)(x + 1) can be 1*3 = 3
Therefore x = 2
Sufficient
Option B

nalinnair · Answer

From the question stem, we get that:
 x  is an integer and  x>0 
Now lets consider the statements:

Statement 1
 (x - 1)  is a prime number
This gives us 2 scenarios:
either  (x - 1)  is an even prime number (i.e. 2) or  (x - 1)  is an odd primer number (i.e. any prime numbers except 2)
If  (x - 1)  is an even prime number, then  x = 3 , an odd integer
If  (x - 1)  is an odd prime number, then  x  has to be an even integer
Therefore, we do not get a definitive yes/no from this statement. This statement is insufficient.
Hence, we can reject options A and D.

Statement 2
 (x^2 - 1)  is a prime number.
This also gives us 2 scenarios:
either  (x^2 - 1)  is an even prime number (i.e. 2) or  (x^2 - 1)  is an odd primer number (i.e. any prime numbers except 2)
If  (x^2 - 1) = 2 ;  x^2 = 3 . This is not possible as we know that  x  is an integer.
Hence, the only possible scenario is that  (x^2 - 1)  is an odd primer number.
This implies  x^2  is an even number and accordingly,  x  is an even number.
This statement is sufficient. We can reject options C and E.

The correct answer is  B .

stonecold · Answer

Excellent Question
Here in statement one prime number can be 2 so X can be even or odd
But in statement two x^2-1 can never be two as it would mean that x is not an integer which is contradictory
so x^2-1 must be odd prime => x^2= odd+odd=even so x must be even
Smash that B

mimajit · Answer

This is how i solved

Statement 1: (x - 1) is a prime number
x = 3 implies (x - 1) = 2. A prime number
x = 4 implies (x - 1) = 3. A prime number.
Hence x can be both even and odd. Insufficient

Statement 2: (x^2 - 1) is a prime number
Only possible value of X^2 -1 being a prime no is 2^2 -1 = 3 Which is not even but an odd prime..

rest when u test values we dont get any value that is prime

Eg : 11^2-1 = 121-1 = 120 Not prime
13^2-1 = 198-1 = 168 not prime

3^2- 1 =8 Not prime ... So only prime no that satisfies the equation is 2^2-1 = 3 Which is an odd prime and we can answer the question with certainty

Alternatively 
(x - 1)(x + 1) is a prime number.
The only possible value of (x - 1)(x + 1) can be 1*3 = 3
Therefore x = 2
Sufficient
Option B

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Is the positive integer x even?

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Is the positive integer x even?

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