Problems on H.C.F and L.C.M
Find the greatest number of four digits which when divided by 10, 15, 21 and 28 leaves 4, 9, 15 and 22 as remainders respectively?
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