This question isn't really that difficult.
Here's how I approached the problem
The question asks when (x^2 + 1)(x + 5) an even number,
We test some numbers to get an idea when it become even.
I used -2,1,0,1,2
The value of expression was (5)(3)=15 when x=-2, (2)(4)=8 when x=-1, (1)(5)=5 when x=0, (2)(6)=12 when x=1,and (5)(7)=35 when x=2
Alternatively, using number properties-> (x^2+1)(x+5) = even when
x^2+1 x+5
Even odd
odd even
even even
You can't have a situation where both are odd
So, we can infer when x is odd the expression is even and when x is even the expression is odd.
So question is "Is x odd? "
1) yes; sufficient
2) each prime factor of x^2 is greater than 7.
Since the expression is x^2 and x is an integer; each prime factor must be repeated atleast 2 times.
We are also told that in the prime factorization of x^2 each prime factor is greater than 7.In other words,each prime factor of x is greater than 7.
Ask yourself the question:When does a number become even? When it has atleast one factor of 2.
2 basically says x can have prime factors such as 11,13,19 but not 2.
No matter what combination you use, you will never get x= even
eg x^2=121 or x=11 =odd (prime factor >7)
take any even number you will find it has atleast one 2 as a factor. For instance x^2= (22)^2 is not a valid choice coz it has 2 as a prime factor
Answer according to me D
