2016^{k} divides
2016!^{2016}.
For how many positive values of the integer
k is the above true?
What would be your answer if the number
2016 was replaced by 2015 in all 2016's appearances?
(In reply to
Paper and Pencil solution, part (a) by Steve Herman)
My solution method is essentially the same, so I'll just share the result to using 2015:
2015 = 5*13*31
So 31 will definitely be the limiting factor.
2015!^2015 is divisible by 31^(67*2015) = 31^135005.
The final answer to the question is 135005/1 =135005
Edited on February 25, 2015, 11:40 am

Posted by Jer
on 20150225 11:04:56 