Data Sufficiency

If x is an integer, is(x^2+)(x+5) an even number?

1) x is an odd number

2) Each prime factor of x^2 is greater than 7.

OAD

If X is an integer, is (x2 + 1)(x + 5) an even number?
(1) x is an odd number.
(2) Each prime factor of x² is greater than 7.

Statement 1: x is odd
If x=1, then (x²+1)(x+5) = (1² + 1)(1 + 5) = 2*6 = 12.
If x=3, then (x²+1)(x+5) = (3² + 1)(3 + 5) = 10*8 = 80.
If x=5, then (x²+1)(x+5) = (5² + 1)(5 + 5) = 26*10 = 260.
The cases above illustrate that -- if x is odd -- then (x²+1)(x+5) = even.
SUFFICIENT.

Statement 2: Each prime factor of x² is greater than 7.
Since each prime factor of x² is greater than 7, each prime factor of X ITSELF must be greater than 7.
Options for x:
11, 13, 17, 19, 23...
Notice that only ODD VALUES FOR X will satisfy statement 2.
As we saw in statement 1, if x is odd, then (x²+1)(x+5) = even.
SUFFICIENT.

The correct answer is D.

Statement 2: Each prime factor of x² is greater than 7.
Since each prime factor of x² is greater than 7, each prime factor of X ITSELF must be greater than 7.
Options for x:
11, 13, 17, 19, 23...
Notice that only ODD VALUES FOR X will satisfy statement 2.
As we saw in statement 1, if x is odd, then (x²+1)(x+5) = even.
SUFFICIENT.

Hi GMATGuruNY ,

Many thanks for your reply.

Just a quick question. In statement 2 its given that x^2 is greater that 7. So will x^2 be 11,13,17...?

Please advise.

Many thanks in advance.

SJ

jain2016 wrote:

Statement 2: Each prime factor of x² is greater than 7.
Since each prime factor of x² is greater than 7, each prime factor of X ITSELF must be greater than 7.
Options for x:
11, 13, 17, 19, 23...
Notice that only ODD VALUES FOR X will satisfy statement 2.
As we saw in statement 1, if x is odd, then (x²+1)(x+5) = even.
SUFFICIENT.

Hi GMATGuruNY ,

Many thanks for your reply.

Just a quick question. In statement 2 its given that x^2 is greater that 7. So will x^2 be 11,13,17...?

Please advise.

Many thanks in advance.

SJ

As noted in my post above:
Since each prime factor of x² is greater than 7, each prime factor of X ITSELF must be greater than 7.
Thus, the least possible values for x are as follows:
11, 13, 17, 19, 23...
Thus, the least possible values for x² are as follows:
11², 13², 17², 23²...
For every value in the blue list, the prime-factorization is composed solely of prime numbers greater than 7.