**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?

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**