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:
Required length = H.C.F. of 495 cm, 900 cm and 1665 cm. 495 = 3² x 5 x 11, 900 = 2² x 3² x 5², 1665 = 3² x 5 x 37. H.C.F. = 32 x 5 = 45. Hence, required length = 45 cm.
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