Problem:
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 kilometers per hour of speed or fraction thereof. (Thus the front of a car traveling 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 meters long and that the cars can travel at any speed, let be the maximum whole number of cars that can pass the photoelectric eye in one hour. Find the quotient when is divided by .
Solution:
Let be the speed of the cars in kilometers per hour. Then the number of car lengths between consecutive cars is (the least integer ), so the distance between consecutive cars is meters. Thus the distance from the front of one car to the front of the next is meters. In one hour each car travels meters. Let the interval from the front of one car to the front of the next be called a gap. Then the number of gaps that pass the eye in one hour is
Let , where . Then
The quantity is positive and approaches zero as increases. Thus , and by taking sufficiently large, can be made as close to as desired. Therefore, at a high enough speed, gaps will pass the eye in one hour. Assuming that at the start of the hour the eye is exactly even with the front of a car, it will be passed by one car in each of the ensuing gaps, and by one additional car at the beginning of the last of a gap. Thus the eye will be passed by cars. The quotient when is divided by is .
Note: With a little more calculation, it can be shown that the minimum speed for which cars pass the eye is . Do not try this at home!
The problems on this page are the property of the MAA's American Mathematics Competitions