A Curious List

Let's start by finding the smallest positive integer $x$ , for which $x | 2016$ and $x^2 \geq 2016$ are true.

We can quite efficiently find that $44^2 < 2016 < 45^2$ , implying that $x > 44$ .

We can also quite efficiently find the prime factorization of $2016$ , which is $2^5 \times 3^2 \times 7$ .

Knowing that $5$ , $23$ and $47$ are primes, we quickly reject $45$ , $46$ and $47$ .

Now since $48 = 2^4 \times 3$ , we find that $x = 48$ because $48 | 2016$ . Also $2016 = 42 \times 48$ .

We know that $x^2 \geq nx$ has $x$ solutions for $n$ when $x$ is a positive integer.

Therefore, the number $2016$ appears first time in the position $1 + 2 + \cdots 47 + 42 = \frac{47 \times 48}{2} \times 42 = 1170$ .

Sum of the digits of $1170$ is $1 + 1 + 7 + 0 = \boxed{9}$ which is the solution.

2016 first appears when we start with the number 48, and $48 * \boxed{42} =2016$ , the boxed number is important as it will be added afterwards.

Now, Starting from the first number:

we have the series like this:

LaTeX: $1, 2, 4, 3, 6, 9, 4, 8, 12, 16, 5,\ldots$

$\boxed{1 number, for 1}, \boxed{2 numbers, for 2}, \boxed{3 numbers, for 3}, \boxed{4 numbers, for 4}, \ldots\$

Now as we see its an AP, we will take till $47^{th}$ term into consideration.

so $\frac{47*48}{2}= \boxed{1128}$

Total terms= $1128+42=1170$ , that means 2016 appears in the $1170^{th}$ position of the series.

So sum of the position= $1+1+7+0=\boxed{9}$