Problems on H.C.F and L.C.M
Let the least number of six digits, which when divided by 4, 6, 10 and 15, leaves in each case the same remainder of 2, be N. The sum of the digits in N is:
The Largest number of four digits is 9999. Required number must be divisible by L.C.M. of 12,15,18,27 i.e. 540. On dividing 9999 by 540,we get 279 as remainder . Required number = (9999-279) = 9720.
EXAMIANS