Suppose the cars travel at $s$ kilometers per hour. The gap is $\lceil s/15 \rceil$ car lengths, so successive fronts are $4\lceil s/15 \rceil + 4$ meters apart, and in one hour a column of $1000s$ meters of traffic passes the eye — that is, $$N = \frac{1000s}{4\lceil s/15 \rceil + 4} = \frac{250s}{\lceil s/15 \rceil + 1}$$ gaps per hour.

For a fixed value $k = \lceil s/15 \rceil,$ the count $N$ is largest at $s = 15k,$ where it equals $\frac{3750k}{k + 1}.$ This is always less than $3750$ but approaches $3750$ as $k$ grows. Although the gap count never reaches $3750,$ the car count can: choose $k$ so large that more than $3749$ gaps pass, and start the hour with a car exactly at the eye. That car, plus one car for each of the $3749$ complete gaps that follow, makes $3750$ cars.

So $M = 3750,$ and the quotient when $M$ is divided by $10$ is $375.$