How many divisors of the combination of numbers? 
Find the number of positive integers that are divisors of at least one of $A=10^{10}, B=15^7, C=18^{11}$

Instead of the PIE formula, I would like to use intuition. 
$10^{10}$ has $121$ divisors, and $15^7$ has $64$ divisors, and $18^{11}$ has $276$ divisors. 
Number of divisors total with no restriction is: $461$. $A,B \to $ there are $5^{0} \to 5^{7} = 8$ divisors. $B, C \to$ there are: $3^{0} \to 3^{7} = 8$ divisors. $A, C \to$ there are $2^{0} \to 2^{10} = 11$ divisors.
So far: $461 - 27 = 434$. I took out: $15$ divisors from $B$, $19$ from $A$, and $19$ from $C$. So in total:
$$434 + 53 = 487$$
But this isnt right.


$10^{10}$ has $121$ divisors, and $15^7$ has $64$ divisors, and $18^{11}$ has $276$ divisors.
$A,B \to $ there are $5^{0} \to 5^{7} = 8$ divisors. $B, C \to$ there are: $3^{0} \to 3^{7} = 8$ divisors. $A, C \to$ there are $2^{0} \to 2^{10} = 11$ divisors.

These are correct. Now $1$ is the only divisor of the three numbers $A,B,C$. So, the number of positive integers that are divisors of at least one of $A,B,C$ is given by
$$(121+64+276)-(8+8+11)+1=435.$$
You may want to see the inclusion–exclusion principle.
