Think of a possibility which would make x!+(x+1) composite (not prime).

Looking at the expression first thing that strikes would be divisibility by 3 (sum of digits should be dvisible by 3)

Now, we have a (x+1) in the expression, so if we had 'x' take values 1 less than multiples of 3 (2,5,8,11 and so on) we will have 'x+1' divisible by 3.

Now again for x! will be divisible by 3 for all x > 3.

Also, adding two numbers divisible by 3, will result in a number again divisible by 3.

Now, look at the problem

Statement 1:

x < 10

We know x!+x+1 cannot be prime for values like 5,8 (1 less than multiple of 3) - Insufficient

Statement 2:

x is even

Again any even number which is 1 less than multiple of 3, would not result in x!+x+1 being prime - Insufficient

Together,

Even together we know value x='8' would not be prime, we get 8!+9 (which would be divisible by 3)
So insufficient.

E IMO.

Good conclusion for solving 8! + 9..

shankar.ashwin wrote:Think of a possibility which would make x!+(x+1) composite (not prime).

Looking at the expression first thing that strikes would be divisibility by 3 (sum of digits should be dvisible by 3)

Now, we have a (x+1) in the expression, so if we had 'x' take values 1 less than multiples of 3 (2,5,8,11 and so on) we will have 'x+1' divisible by 3.

Now again for x! will be divisible by 3 for all x > 3.

Also, adding two numbers divisible by 3, will result in a number again divisible by 3.

Now, look at the problem

Statement 1:

x < 10

We know x!+x+1 cannot be prime for values like 5,8 (1 less than multiple of 3) - Insufficient

Statement 2:

x is even

Again any even number which is 1 less than multiple of 3, would not result in x!+x+1 being prime - Insufficient

Together,

Even together we know value x='8' would not be prime, we get 8!+9 (which would be divisible by 3)
So insufficient.

E IMO.

but for statement 2 we haven't checked if there is any number which results in x! + (x+1) being prime. Until we check such a case how can we be sure that statement 2 is insuffcient

Its obviously prime for numbers such as x=1 (3) or x =2 (5). In some cases the expression is prime while in some others its not. So B alone will not be sufficient.

vishal.pathak wrote: but for statement 2 we haven't checked if there is any number which results in x! + (x+1) being prime. Until we check such a case how can we be sure that statement 2 is insuffcient