2016k 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?
Well, 2016 = 2^5 * 3^2 * 7.
I suspect that 7 will be the limiting factor.
Let's count what power of 7 divides into 2016!.
2016/7 = 288, so 288 numbers in 2016! are divisible by 7.
288/7 = 41 (rounding down), so 41 of the 288 are divisible by 7^2
41/7 = 5 (rounding down), so 5 of the 41 are divisible by 7^3.
288+41+5 = 334, so 2016! is divisible by 7^334.
And 2016!^2016 is divisible by 7^(334*2016) = 7^673344.
7 is the limiting factor if :
powers of 3 in 2016! are greater than 2*334 = 668
and powers of 2 in 2016! are greater than 5*334 = 1670
Well, that is in fact the case,
because 2016/3 = 672 are divisible by 3. That is enough for us to disregard 3.
As for 2,
1008 factors of 2016 are divisible by 2
504 factors are divisible by 4
252 are divisible by 8,
and we need go no further, because 1008 + 504 + 252 > 1670
The final answer to the question is 673,344/1 = 673,344
Paper and Pencil solution, part (a)
