- anirudhbhalotia
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Statement 1: k > 4!
k may be prime (29, 31 etc) or may be composite (25, 26 etc.)
Not sufficient.
Statement 2: (13! + 2) ≤ k ≤ (13! + 13)
Each possible value of k is composite integer. Take few for example,
- (1) k = (13! + 2) = (13*12*11*10*9*8*7*6*5*4*3*2*1 + 2) = 2*(13*12*11*10*9*8*7*6*5*4*3*1 + 1) => Multiple of 2
(2) k = (13! + 3) = (13*12*11*10*9*8*7*6*5*4*3*2*1 + 3) = 3*(13*12*11*10*9*8*7*6*5*4*2*1 + 1) => Multiple of 3
(3) k = (13! + 4) = (13*12*11*10*9*8*7*6*5*4*3*2*1 + 4) = 4*(13*12*11*10*9*8*7*6*5*3*2*1 + 1) => Multiple of 4
(4) Same for 5, 6, 7, 8, 9, 10, 11, 12 and 13
The correct answer is B.














