TITLE: PIE Problem with divisors
QUESTION [0 upvotes]: Find the number of positive integers that are divisors of at least one of $10^{10},15^7,18^{11}$.

Let $n(A)$ be the number of positive integers that divide $10^{10}$ let $n(B)$ be the number of positive integers that divide $15^7$. Let $n(C)$ be the number of positive integers that divide $18^{11}$. 
$$n(A) \to 10^{10} = (2 \cdot 5)^{10} = 2^{10} \cdot 5^{10} \implies \text{121 divisors.} $$
$$n(B) \to 15^{7} = 5^{7} \cdot 3^{7} \implies \text{64 divisors.}$$
$$n(C) \to 18^{11} = 2^{11} \cdot 3^{22} \implies \text{376 divisors.}$$
$$n(A \cup B \cup C) = n(A) + n(B) + n(C) - n(A \cap B \cap C)$$
Calculating $n(A \cap B \cap C)$ is tough, I think I get this incorrectly. $1$ divides all numbers. That's all. 
$$n(A \cup B \cup C) = n(A) + n(B) + n(C) - n(A \cap B \cap C) = 561 - 1 = 560$$

REPLY [3 votes]: The problem is that your PIE formula is wrong; it should be
$$\begin{align*}
n(A\cup B\cup C)&=n(A)+n(B)+n(C)\\
&\quad-\big(n(A\cap B)+n(A\cap C)+n(B\cap C)\big)\\
&\quad+n(A\cap B\cap C)\;.
\end{align*}$$
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