Let D(x) denote the total number of divisors of x, including 1 and itself.

Thus, taking m = 10^40 = (5^40)*(2^40), and:
n = 20^30 = (2^60)*(5^30), we have:

p = GCD(m, n) = (2^40)*(5^30)

Thus, the required number of divisors (v) dividing at least one of 10^40 and 20^30 is given by:

V = D(m)+ D(n) - D(p)
= 41^2 + 61*31 - 41*31
= 2301