Question 1210186
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Find the number of positive integers that are divisors of at least one of 6^{6}, 10^{10}, 15^{15}, and 30^{30}.
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<pre>
Notice that all three numbers  {{{6^6}}},  {{{10^10}}}  and  {{{15^15}}}  are divisors of the number  {{{30^30}}}.


Therefore, you can simplify the problem's formulation: you ONLY need to find the number of divisors of  {{{30^30}}}.


The prime decomposition for 30 is  30 = 2*3*5.


So,  {{{30^30}}} = {{{2^30*3^30*5^30}}}.


For this number, its divisors are all  

    1,  2,  {{{2^2}}}, . . . ,  {{{2^30}}}   (31 divisor),

    1,  3,  {{{3^2}}}, . . . ,  {{{3^30}}}   (31 divisor),

    1,  5,  {{{5^2}}}, . . . ,  {{{5^30}}}   (31 divisor)


and all of their products.


The number of all their distinct products is  31*31*31 = {{{31^3}}} = 29791.


It is the <U>ANSWER</U> to the problem's question.
</pre>

Solved.


I wrote this my post after @CPhill to provide the most simple straightforward solution/explanation
and to remove all unnecessary complications from the post by @CPhill.



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Edwin, I think it was very good from your side to make this editing in your post
and to exclude/ (to remove) your unjustified insinuations against me from the original version.