The answer is 44.

**
This section requires Javascript.
**

You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.

We have $L = 2^{10} \cdot 3^6 \cdot 7^3 \cdot P,$ where $P$ is a product of other prime powers. Then $L/2016 = 2^5 \cdot 3^4 \cdot 7^2 \cdot P.$ So we are looking for numbers which are divisible by $2^6,$ or $3^5,$ or $7^3.$ No number $\le 2017$ is divisible by two of these simultaneously, so we can count each one separately.

There are $\lfloor 2017 / 2^6 \rfloor = 31$ numbers divisible by $2^6,$ $\lfloor 2017 / 3^5 \rfloor = 8$ numbers divisible by $3^5,$ and $\lfloor 2017 / 7^3 \rfloor = 5$ numbers divisible by $7^3,$ so the answer is $31 + 8 + 5 = \fbox{44}.$