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?
2016 is 2^5 * 3^2 * 7^1.
2016! has 2 as a factor 2010 times; 3 as a factor 1004 times; and 7, 334 times.
(the above was determined by
n = 2016
For i = 1 To n
ii = i
While ii Mod 2 = 0
fct2 = fct2 + 1
ii = ii / 2
Wend
While ii Mod 3 = 0
fct3 = fct3 + 1
ii = ii / 3
Wend
While ii Mod 7 = 0
fct7 = fct7 + 1
ii = ii / 7
Wend
Next
Text1.Text = Text1.Text & fct2 & Str(fct3) & Str(fct7) & crlf
)
That factorial raised to the 2016 thus has these factors 4052160, 2024064 and 673344 times respectively.
The powerof2 requirements for 2016 would be satisfied 4052160/5 = 810432 times; powerof3, 2024064/2 = 1012032 times and powerof7, 673344 times. The limiting factor is then the power of 7: k can be as high as 673,344.
2015 has as factors 5, 13 and 31, each once.
The counts for these factors in 2015! are 502, 166 and 67 respectively, so the 31asafactor will be the limiting one. When raised to the 2015, this comes to 67*2015= 135,005, for the highest allowable k.

Posted by Charlie
on 20150225 10:50:12 