Will $3^r$ divides $n$ ?

$2017-\left\lfloor \frac{2017}{3}\right\rfloor\Rightarrow 1345$

Doing the problem per the definition directly:

$\text{Length}\left[\text{Union}\left[\text{Table}\left[\left(n\to \frac{n}{3^{\text{IntegerExponent}[n,3]}}\right)(n),\{n,2017\}\right]\right]\right] \Rightarrow 1345$

Reading the Wolfram Mathematica expression into English: the size of the set (Length) reduced from the list (Union) that results from applying the function $\left(n\to \frac{n}{3^{\text{IntegerExponent}[n,3]}}\right)$ to the integers ( $n$ ) from 1 to 2017 in turn gives $1345$ . The function evaluation $\text{IntegerExponent}[n,3]$ returns the number of factors of 3 in $n$ .

This problem is equivalent to asking how many integers between 1 and 2017, inclusive, do not contain a factor of 3, as the residual after removing the factors of 3 is still a positive integer and will be included in that manner:

$\text{Tally}[\text{Table}[\text{IntegerExponent}[n,3],\{n,2017\}]]\Rightarrow \left( \begin{array}{cc} 0 & 1345 \\ 1 & 448 \\ 2 & 150 \\ 3 & 50 \\ 4 & 16 \\ 5 & 6 \\ 6 & 2 \\ \end{array} \right)$

The number is 2017- the number of multiples of 3, which is 672. So the number is 2017-672=1345