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.
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.
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