AIME I #12 2008

[stupidityismygam]

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 [unparseable or potentially dangerous latex formula] kilometers per hour of speed or fraction thereof. (Thus the front of a car traveling [unparseable or potentially dangerous latex formula] kilometers per hour will be four car lengths behind the back of the car in front of it.) A photoelectric eye the side of the road counts the number of cars that pass in one hour. Assuming that each car is [unparseable or potentially dangerous latex formula] meters long and that the cars can travel at any speed, let [unparseable or potentially dangerous latex formula] be the maximum whole number of cars that can pass the photoelectric eye in one hour. Find the quotient when [unparseable or potentially dangerous latex formula] is divided by [unparseable or potentially dangerous latex formula]

[Dragon of Light]

Consider the car and the space behind it as one "block".

Count how many "blocks" pass the Eye.
Each block has length [unparseable or potentially dangerous latex formula] km, where x is the speed in km/hr.

We can assume the speed is 15k, where k is an integer, because any speed between 15k and 15(k-1) produces blocks of the same size, yet more cars pass the Eye when the speed is faster.

The number of blocks that pass the Eye is [unparseable or potentially dangerous latex formula].

Substituting x=15k, we get [unparseable or potentially dangerous latex formula].

The limit as k approaches infinity of the equation is just [unparseable or potentially dangerous latex formula] or 3750.

It is true that the speed cannot be infinity. However, 3750 represents the number of blocks. It is possible for 3750 cars to pass the Eye at a finite speed, even though the space between the 3750th and 3751th car may not fully pass the Eye.

Thus our answer is 3750/10=375.