RRB RPF RPSF SI and CONS Model Papers 5 with solutions

(4) LCM of 6, 8 and 12 = 24
Therefore, the required number
= 24k - 1 (1 = 6 - 5 = 8 - 7= 12 - 11)
24k - 1 = 13 - 2k - (2k + 1)
The least value of k for which 13 ×2k × (2k + 1) is divisible by 13 is 6.
Therefore, the required number = 24 × 6 × 1 = 143
Sum of the digits = 1 + 4 + 3 = 8