First consider (1) alone.
x<10.
For x = 1, 2, 3, 4 the values of x! + (x + 1) are 2, 5, 10, 29 respectively.
Out of these, 3 are prime and 1 is not prime.
Or (1) alone is not sufficient.
Next consider (2) alone.
For x = 2, 4, 6, 8 the values of x! + (x + 1) are 5, 29, 727, 8!+9.
The first 3 are prime but 8! + 9 will not be a prime because it will be divisible by 9.
Again nothing definite can be said.
So even (2) alone is not sufficient.
On combining we can check again with same values of 2, 4, 6 and 8 and get the same result that nothing definite can be said .
The correct answer is E.
prime number
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Source: Beat The GMAT — Data Sufficiency |
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Rahul@gurome
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Rahul Lakhani
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Laura GMAT Tutor
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Rahul's explanation is great. 
But I wanted to throw in something else...
He said that "8! + 9" will be divisible by 9, and I just wanted to clarify why that was, in case anyone was wondering.
First, of course, 8! is (8)(7)(6)(5)(4)(3)(2)(1). Will 8! be a multiple of 9? Absolutely! It has a 6 and a 3, meaning that it has two 3s in its prime factor list. Since 8! must be a multiple of 9, 8!+9 must also be a multiple of 9. So it can't be prime.
But I wanted to throw in something else...He said that "8! + 9" will be divisible by 9, and I just wanted to clarify why that was, in case anyone was wondering.
First, of course, 8! is (8)(7)(6)(5)(4)(3)(2)(1). Will 8! be a multiple of 9? Absolutely! It has a 6 and a 3, meaning that it has two 3s in its prime factor list. Since 8! must be a multiple of 9, 8!+9 must also be a multiple of 9. So it can't be prime.
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rishab1988
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