Question

On a long straight stretch of one-way single-lane highway, cars all travel at the same speed and all obey the safety rule: the distance from the back of the car ahead to the front of the car behind is exactly one car length for each $15$ kilometers per hour of speed or fraction thereof. (Thus the front of a car traveling $52$ kilometers per hour will be four car lengths behind the back of the car in front of it.)

A photoelectric eye by the side of the road counts the number of cars that pass in one hour. Assuming that each car is $4$ meters long and that the cars can travel at any speed, let $M$ be the maximum whole number of cars that can pass the photoelectric eye in one hour. Find the quotient when $M$ is divided by $10.$

Answer

Answer: 375

Concepts:floor and ceiling functions optimization rate

Difficulty rating: 2920

Solution:

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 $375.$

2008 AIME I Problem 12

Answer: 375

Difficulty rating: 2920

Solution:

Problem 12 in Other Years