# Let $X, Y,$ and $Z$ be the sets of positive divisors of $10^{60}, 20^{50}$ and $30^{40}$ respectively. Find $n (X\cup Y\cup Z)$.

Let $$X, Y,$$ and $$Z$$ be the sets of positive divisors of $$10^{60}, 20^{50}$$ and $$30^{40}$$ respectively. Find $$n (X\cup Y\cup Z)$$.

I've trying to solve this question since long time but I am unable to do so. I have tried to use Venn diagrams but such approaches did not help me. I am not good at combinatorics so therefore I am seeking help? Would someone please help me to solve this question?

Thanks for help!

• What is $n(AUBUC)$? Do you mean the cardinality of the set $X\cup Y\cup Z$ ? – WaveX Nov 29 '19 at 16:46
• Yes. But I don't know how to type that. – Shashwat1337 Nov 29 '19 at 16:49
• "How to type that": math.meta.stackexchange.com/questions/1773/… – David K Nov 29 '19 at 16:51
• What does the title have to do with the question in the text? – joriki Nov 29 '19 at 17:28
• Please write a title that is specific to the problem you wish to solve. – N. F. Taussig Nov 29 '19 at 17:31

Note that $$10^{60} = 2^{60} \times 5^{60}$$ , $$20^{50} = 2^{100} \times 5^{50}$$ and $$30^{40} = 2^{40} \times 3^{40}\times 5^{40}$$.
Divisors of the form $$2^{i} \times 5^{j}$$ are shown in the diagram below.
The other factors have the form $$2^{i} \times 3^{j} \times 5^{k}$$ where $$i=0,1,\cdots,40$$ ,$$j=1,2\cdots,40$$ and $$k=0,1,\cdots,40$$.
Now we just need to do the arithmetic $$610+3111+2040+ 41 \times 40 \times41=?$$.