GMAT Math

If x is an integer, is (x^2 + 1)(x + 5) an even number?

1. X is an odd number

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

OA: D

I get part 1 but why 2?

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

Hey haidgmat:

Good question on statement 2 - by saying that "each prime factor is >7", the statement is really saying "x is odd" and just repeating statement 1 but in a less-convenient way.

Here's why - the ONLY way for a (nonzero) integer to be even is to have 2 as one of its prime factors. And statement 2 rules that out by saying that it doesn't have any prime factors less than 7 (and 2, as we know, is a prime number less than 7).

As rishab1988 posted, this comes down to Odd vs. Even, so almost immediately you should start thinking about "divisible by 2" as the rule for even. Because both statements 1 and 2 prohibit an even number for x, then we know that:

(Odd^2 + 1) is going to be Even.
(Odd + 5) is going to be Even

Even * Even = Even

And therefore the number will be even.