You should notice that f is not multiplicative, as f(1) = (1 + 1) = 2, so if it was multiplicative, f(k) = f(k) * f(1) = 2f(k) (Contradiction!). Therefore, that is not the approach.

For solving the problem, I'm thinking the following fact should be useful: Let's remember that if we enumerate prime numbers p₁ = 2, p₂ = 3, ..., a = p₁^α₁p₂^α₂..., and b = p₁^β₁p₂^β₂...,then LCM(a, b) = p₁^{max(α₁, β₁)}p₂^{max(α₂, β₂)}.... From here, we can draw an important observation: if LCM(a, b) = n, and the maximum power of p_i that divides n is ζ_i, then either the power of p_i in a is ζ_i or the power of p_i in b is ζ_i (or both). As the number of distinct prime factors is at most 15, there will be no more than 2¹⁵ ways of assigning the ζ_i to each member of the pair. Then, you can calculate the sums more or less easily.