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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!

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

enter image description here

And the number of factors is indicated in each rectangle.

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 $610+3111+2040+ 41 \times 40 \times41=?$.

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  • $\begingroup$ Thanks a lot Donald $\endgroup$ – Shashwat1337 Nov 30 '19 at 10:39

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