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.
Since the numbers are co-prime, they contain only 1 as the common factor.Also, the given two products have the middle number in common.So, middle number = H.C.F. of 551 and 1073 = 29;First number = 551/29 = 19; Third number = 1073/29 = 37.Required sum = (19 + 29 + 37) = 85.
EXAMIANS