<s>[INST] 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.
 [/INST] 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.
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