Veritas DS - reasoning reqd

[karthikpandian19]

[Anurag@Gurome]

Given: If an integer g have a factor f such that 1 < f < g, then g is a composite number, otherwise f is prime. Thus the question simply asks whether g is prime or composite.

Statement 1: g > 3! or g > 6
g may be prime (11, 13 etc) or may be composite (12, 15 etc.) ; NOT sufficient.

Statement 2: (11! + 11) ≥ g ≥ (11! + 2)
Each possible value of g is composite integer. Take few for example,
(1) g = (11! + 2) = (11*10*9*8*7*6*5*4*3*2*1 + 2) = 2*(11*10*9*8*7*6*5*4*3*1 + 1) => Multiple of 2

(2) g = (11! + 3) = (11*10*9*8*7*6*5*4*3*2*1 + 3) = 3*(11*10*9*8*7*6*5*4*2*1 + 1) => Multiple of 3

(3) g = (11! + 4) = (11*10*9*8*7*6*5*4*3*2*1 + 4) = 4*(11*10*9*8*7*6*5*3*2*1 + 1) => Multiple of 4

(4) Same for 5, 6, 7, 8, 9, 10, and 11; SUFFICIENT.

The correct answer is B.

[aneesh.kg]

Statement(1) is obviously INSUFFICIENT.

Statement(2):
Let
g = 11! + k
where
2 =< k =< 11
If this is the case then 'k' is already contained inside 11! because 11! is a product of all integers from 1 to 11, through 'k'.
g = 11*...(k)...*1 + k = k*(11*... ..*1 + 1) = k*Integer = A multiple of k

As we saw, 'g' is a multiple of all numbers from 1 to n.
or
1 =< f =< 11
SUFFICIENT

[spoiler](B)[/spoiler] is the answer