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 :
L.C.M. of 5,6,7,8 = 840. Required number is of the form 840k + 3 Least value of k for which (840k + 3) is divisible by 9 is k = 2. Required number = (840 X 2 + 3)=1683
EXAMIANS