Starting with Numbers..:P

From the way the set is defined it's clearly that any element should be a divisor of at least one of the three numbers.

If $A = {D}_{{10}^{20}}$ , $B = {D}_{{15}^{10}}$ , $C = {D}_{{24}^{20}}$ we need to calculate $|A \cup B \cup C|$ .

We should know that if $n = {{p}_{1}}^{{k}_{1}}{{p}_{2}}^{{k}_{2}}...{{p}_{m}}^{{k}_{m}}$ is a prime factorization of $n$ (when $n > 1$ ), then the number of divisors of $n$ is:

$|{D}_{n}| =(1 + {k}_{1}) (1 + {k}_{2}) ... (1 + {k}_{m})$

(eg. If $n=60={2}^{2}{3}^{1}{5}^{1}$ then 60 should have ${D}_{60} = (1 + 2)(1 + 1)(1 + 1)=12$ divisors, which is true because ${D}_{60} = \{1,2,3,4,5,6,10,12,15,20,30,60\}$ )

The principal of inclusion and exclusion says that

$|A\cup B \cup C| =|A| + |B| + |C| - |A\cap B| -|A \cap C| - |B \cap C| + |A \cap B \cap C|$

${10}^{20} = {2}^{20} {5}^{20}$ then $|A|=(1+20)(1+20)=441$

${15}^{10} = {3}^{10}{5}^{10}$ then $|B|=(1+10)(1+10)=121$

${25}^{15} = {2}^{45} {3}^{15}$ then $|A|=(1+45)(1+15)=736$

$|A \cap B| = {D}_{{5}^{10}} = 1 + 10 = 11$

$|A \cap C| = {D}_{{2}^{20}} = 1 + 20 = 21$

$|B \cap C| = {D}_{{3}^{10}}= 1 + 10 = 11$

$|A \cap B \ cap C|= {D}_{1} = 1$

Plug everything in the formula of $|A \cup B \cup C|$ .

answer is probably one of 1024 ,1256 or 1266 as per the choices given where I found this question or may be none of these..:P