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.
 A: 
$10^{10}$ 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.
A: Your $435$ in the comment is correct. As stated in the comments, only $1$ divides all three numbers. You counted it three times, once for each number, then subtracted it three times, once for each pair, so now you need to add it back in once.
A: Each divisor is of the form $2^i3^j5^k$, where the exponents are nonnegative.
If $i=0$ and $j=0$, then $0\le k \le 10$, which accounts for $11$ possibilities.
If $i=0$, $j>0$, and $k=0$, then $1\le j \le 22$, which accounts for $22$ possibilities.
If $i=0$, $j>0$, and $k>0$, then $1\le j,k \le 7$, which accounts for $49$ possibilities.
If $i>0$, $j=0$, and $k=0$, then $1\le i\le11$, for $11$ more.
If $i>0$, $j=0$, and $k>0$, then $1\le i\le10$ and $1\le k \le10$, for $100$ more.
If $i>0$ and $j>0$, then $k=0$, $1\le i\le11$, and $1\le j \le22$, for $242$ more.
This analysis by cases is exhaustive, and there are $11+22+49+11+100+242=435$ factors.
A: I get a rather different, smaller answer: 416. I think the answers above are counting some divisors twice, thus inflating the cardinality of the desired set. My reasoning is as follows:
N1 = 10^10 = 2^10 * 5^10; divisors = 11*11 = 121
N2 = 15^7  = 3^7  * 5^7 ; divisors = 8*8   =  64
N3 = 18^11 = 2^11 * 3^22; divisors = 12*23 = 276
No number is a divisor of N1 & N2 & N3, except 1.
All of {divisors of at least one of N1, N2, N3} fall into one of these 7 disjoint categories:
Divisors of N1 only: 2s and 5s: 100 (2^n * 3^0 * 5^m)
Divisors of N2 only: 3s and 5s:  49 (2^0 * 3^n * 5^m)
Divisors of N3 only: 2s and 3s: 242 (2^n * 3^m * 5^0)
Divisors of N1 & N2: 5s only:     7 (2^0 * 3^0 * 5^n)
Divisors of N2 & N3: 3s only:     7 (2^0 * 3^n * 5^0)
Divisors of N3 & N1: 2s only:    10 (2^n * 3^0 * 5^0)
Divisors of N1 & N2 & N3:         1 (2^0 * 3^0 * 5^0)
(All n, m above are NOT allowed to be 0; otherwise, they duplicate other categories.)
Hence the set {divisors of at least one of N1, N2, N3} 
has exactly 416 members.
I think the previous answers are too large because they allowed 0 exponents on some prime factors, resulting in numbers which belong to more than one of my 7 disjoint categories, hence erroneously inflating "total members" of the set of {divisors of at least one of N1, N2, N3}. 
