Problems on H.C.F and L.C.M
Let the leas 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 6, 9, 15 and 18 is 90.Let required number be 90k + 4, which is multiple of 7.Least value of k for which (90k + 4) is divisible by 7 is k = 4.Required number = (90 x 4) + 4 = 364.
EXAMIANS