RRB NTPC Mains or Stage 2 Model Paper 7 with solutions

(a) The LCM of 6, 7, 8 , 9 and 10 = 2520
The greatest number of 6 digits = 999999
Dividing 999999 by 2520, we get 2079 as remainder. Hence, the 6-digit number divisible by 2520, is (999999 – 2079), or 997920.
Since 6 – 4 = 2, 7 – 5 = 2, 8 – 6 = 2, 9– 7 = 2, 10– 8 = 2, the remainder in each case is less than the divisor by 2.
∴the required number = 997920 – 2 = 997918