SSC CPO SI and ASI Model Paper 10

(4) 27 = 3 × 3 × 3
39 = 3 × 13
48 = 2 × 2 × 2 × 2 × 3
The least positive integer divisibleby 27, 39 and 48 is the LCM of thesenumbers = 2 × 2 × 2 × 2 × 3 × 3× 3 × 13 = 5116
Therefore, the required squarenumber = 2 × 2 × 2 × 2 × 3 × 3 × 3× 3 × 13 × 13 = 219024