Let $A$, $B$ and $C$ be the sets of the divisors of $10^{10}$, $12^{12}$ and $15^{15}$, respectively. Then the problem actually asks for $n(A\cup B\cup C)$. In this solution, $(a,b)$ denotes the GCD of $a$ and $b$.

So, $A\cap B$ is the set of common divisors of $10^{10}$ and $12^{12}$. Recall that $d$ is a common divisor of $a$ and $b$ if and only if $d\mid (a,b)$. Hence, $A\cap B$ is necessarily the set of the divisors of $(10^{10}, 12^{12})=2^{10}$. Similarly, $B\cap C$ and $A\cap C$ are the sets of the divisors of $3^{12}$ and $5^{10}$, respectively. Finally, $A\cap B\cap C = \{1\}$ as $(10^{10}, 12^{12}, 15^{15})=1$. From these information, we now calculate the number of elements of all these sets:

$n(A) = 11\times11=121$

$n(B) = 25\times13=325 $

$n(C) = 16\times16=256  $

$n(A\cap B) = 11  $

$n(B\cap C) = 13 $

$n(A\cap C) = 11  $

$n(A\cap B\cap C) = 1 $

Therefore, the answer is

$n(A\cup B\cup C) = n(A)+n(B)+n(C)-n(A\cap B)-n(B\cap C)-n(A\cap C)+n(A\cap B\cap C)$

$= 121+325+256-11-13-11+1$

$= \boxed{668}$