How To Crack The 25-Car Puzzle? - An illustration showing a fancy-looking race car that is being multiplied 25 times. Around this fancy illustration, the following text is written: The 25-Car Puzzle.

The 25-Car Puzzle features a simple setting but can be deceptively challenging. I recently came across this puzzle from Christopher D. Long and found it very engaging. So, I thought I would write about it.

Our puzzle begins with a paddock of 25 race cars. You are to figure out which of these are the fastest three. To do this, you need to stick to the following rules:

1. At any given time, only five cars may race at the most.

2. You get to know ONLY the order of the finish and NOT the timings.

3. You are to figure out the fastest three race cars in the fewest number of races!

Do you think you can crack this puzzle? Remember, if your solution appears straightforward, it is probably not the right solution. This is a tricky puzzle.

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Spoiler Alert

Beyond this section, I will be explicitly discussing the solution to the puzzle. If you wish to solve it on your own, I suggest that you pause reading this essay now.

Once you are done with your attempt, you may continue reading the essay and compare approaches. All the best!

My Initial Approach — A Roundabout Trip

When I started solving this puzzle, I started out with the most straightforward approach. We begin by grouping five lots of five race cars. Then, they get to race in five heats.

At the end of these five races, we would have five lots of [1, 2, 3] finishers. That is, the five races would eliminate 10 cars in total, and we would still have 15 cars in contention.

We could then group the remaining 15 cars into three lots of five each and run a second round of heats. This would eliminate 6 more cars, resulting in 9 left in contention.

Continuing the same approach, we could conduct two more races, with one lot of five cars and another lot of four cars. At the end of this third heats-round, we would have 6 cars left.

How To Crack The 25-Car Puzzle? — An initial approach shows that we can figure out the fastes three cars at the end of 12 races.
The Initial Approach — Illustrative art created by the author

In order to figure out the fastest three from this point on, we could hold 2 of these cars in the paddock, and let 4 of them race. This would eliminate one more car. Finally, we could race the remaining five cars to figure out the fastest three.

This approach, albeit straightforward, costs us 12 races in total. As I had mentioned previously, we could do better than this; much better than this!

How to Crack the 25-Car Puzzle?

In this refined approach, we start out the same way we did before: we run five heats with five lots of five race cars. At the end of this round, we would have [5 * 3 = 15] cars left in total.

But unlike before, we could pick the top car out of each heat (the number 1s) and conduct race no. 6 between them. This would tell us which car is the fastest straight away.

The second and third cars in this race could be the overall second and third. But we wouldn’t know that for sure yet. So, we need to pick the number 2s (the cars that finished second) in the first five heats races and run a new race number 7 between them.

Since we have already established to overall fastest car, we are only looking for two more places (2 and 3). Consequently, it suffices if we select the first two cars from race number 7.

How To Crack The 25-Car Puzzle? — A more refined approach shows that we can figure out the fastest three race cars at the end of 9 races.
A Refined Approach — Illustrative art created by the author

Between the 2nd and 3rd from race 6 and first and second from race 7, we surely have the second fastest car. But there is still an outside possibility for a third fastest car.

To cover this, we need to run race 8 between all third fastest cars from races 1 to 5. Finally, we would be left with five cars (2nd and 3rd from race-6, 1st and 2nd from race-7, and 1st from race 8). A final race-9 between these cars would give us the overall second and third fastest race cars.

This approach would get us the result in 9 races; 3 lesser than we started out with. But believe it or not, there is an even more efficient method!

How to Crack the 25-Car Puzzle? — For Real This Time!

Tobe frank with you, I did not figure this one out (as much as I wish I did). We start out the same way as we did before with five heats between five lots of five race cars each.

At the end of this round, we would have five lots of [1, 2, 3] positions respectively. Next, we take the first position cars from each of these races, and run race-6 between them to establish the fastest car. So far, so good.

Now comes the really counter-intuitive bit. The second place car from race-6 is faster than all the other first placed cars other than the one that finished before it.

This means that this car is faster than all the other groups, EXCEPT the group that the overall fastest car raced in.

So, the second fastest car HAS to be one of the following cars: the second car from race-6 or the second from group-1 (where the fastest car raced originally).

Applying a similar logic, the third fastest car from race-6 is faster than all the groups except the groups that the first two (from race-6) belonged to.

How To Crack The 25-Car Puzzle? — The final solution reveals that we can identify the fastest three race cars in just 7 races.
The Final Solution to the 25-Car Puzzle — Illustrative art created by the author

Consequently, the only other contenders for the overall third fastest car are the following: the third fastest car from race-6, the third fastest car from group-1 (where the fastest car raced originally) and the second fastest car from group-2 (where the second fastest car from race-6 raced).

To figure out the second and third fastest cars, we need only run one more race with the following contenders:

1. The second car from race-6.

2. The second car from group-1 (where the fastest car raced originally).

3. The third car from race-6.

4. The third car from group-1 (where the fastest car raced originally).

5. The second car from group-2 (where the second fastest car from race-6 raced).

At the end of this race, we will have figured out the fastest three cars. This approach would cost us only 7 races as opposed to the 12 we originally started out with.

Conclusion

Having come this far, I still do not know if there are any better possible solutions. So, I leave that part open to the reader/puzzle enthusiast.

If you enjoyed cracking this puzzle, then keep an eye on this space for more engaging puzzles in the future!

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Further reading that might interest you: 

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