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.
Last edited by
shankar.ashwin on Fri Nov 18, 2011 10:54 pm, edited 1 time in total.
