Problems on H.C.F and L.C.M
Find the least number which when divided by 6, 7, 8, 9 and 10 leaves 1, 2, 3, 4 and 5 as remainders respectively, but when divided by 19 leaves no remainder?
L.C.M. of 5, 6, 7, 8 = 840.Required number is of the form 840k + 3Least value of k for which (840k + 3) is divisible by 9 is k = 2.Required number = (840 x 2 + 3) = 1683.
Let the numbers be 37a and 37b.Then, 37a x 37b = 4107 ab = 3.Now, co-primes with product 3 are (1, 3).So, the required numbers are (37 x 1, 37 x 3) i.e., (37, 111). Greater number = 111.
EXAMIANS