Mathematics question: What is the correct answer?

Without giving you the answer, I can tell you how I’d approach the problem that you presented, namely: “How can I determine the three fastest horses out of 25, when I can only race five at a time together?”

In the first place, because you’ve asked for “the three fastest”, then it’s not just “which ones can win their heats?” (when I race five at a time), because some horses, like some humans, won’t necessarily run their fastest in a heat if all they have to do is beat four other relatively slow competitors to win. You only race as fast as you need to in order to win, which won’t tell you much about “who is fastest overall”.

So I’d definitely race five heats of five horses each, but I’d time both the winners and the second-place finishers (at least), because if you have a heat made up of your five fastest horses and only take the winner from that heat, plus the winners from your other four heats, and have a final race, then you may have unknowingly left out the second and third fastest horses from the finals.

The races have to be timed, in other words, so that you know (at least within heats) which are the fastest horses. You may not necessarily want to have a final composed of only the winners of the five heats.

I think you mean “how many heats will be required and not “how many number of horses will be required”

As with any problem involving confound variables (such as the effort each horse puts in to the race, or how they are running on that particular day) running more races will give you a more accurate answer as their times average out. For the sake of the question though, lets assume a faster horse will always run faster.

Given the question asks for the minimum number of races, all you need to do is time each horse in 5 races of 5 horses each. If you do not have a timing device, then you will need to run each horse against suitable opponents without having a quantitative measure of its speed. To prove a horse is not one of the 3 fastest, it must be beaten by 3 horses over the course of the tournament. 25 enter, and if you discard the two slowest from each heat, you are left with 15. From here it gets tricky. A horse that places first must win to progress. A horse that came second must come second again, or else first.
The question asks for the minimum number, so lets assume favourable results. However bear in mind that results do affect the number of heats that will have to be run. After the group of 15 do their three heats, only 3 horses will have won twice. However one horse in the third group will have had an easy run, having competed against second tier horses. This one will need to run again against those that came first in round one and second in round two to decide which is the third fastest.
This leaves us with three rounds, with a total of nine races.

Arrange 5 races with 5 horses in each race. The last two horses in each race can be eliminated as there are three faster horses than each of them.

Next, race the 5 fastest horses from original 5 races. The fastest horse in this race will be the fastest of the 25. It is also possible to eliminate more horses based on this result as the original races can be ranked now we know how the winners compare and any horse can be eliminated if there are 3 horses faster than it.

This leaves 6 horses, the fastest of all and five others. In the 7th and final race we race these 5 and the first and second horses are the ones we want. So 7 races are required.

But if you remove a horse who’s slower on the 1st run of 5 horses… that horse can be faster than then other 20 remaining so this can’t apply .

@Hibernate It could be faster than the other 20, that is correct but as it is proven to be slower than three other horses it cannot be in the fastest 3 and so it is safe to eliminate it.

Yes. The answer is 7 races.

I have solved in this way.

Race all 25 horses in groups of 5. Call the groups A, B, C, D and E. Call the winners from each of these races A1, B1, C1, D1, and E1. That’s 5 races so far, and any horse who came in 4th or 5th in any race is out.

Now, race A1, B1, C1, D1, and E1 together. That’s 6 races. Let’s just assume that A1 wins, B1 is second, and C1 is third. That means:

(1) A1 is definitely the fastest horse overall.
(2) Any horse from group D or group E is out.
(3) Any horse from group C, other than C1, is definitely out—they lost to C1, who was third in the winners’ race; thus for example, C2 is fourth fastest at best.
(4) B3 is out. He came in third to B1, thus he’s at best 4th fastest overall.

Thus, the only remaining horses vying for contention are A1, A2, A3, B1, B2, and C1. We know A1 is definitely the fastest. Thus race A2, A3, B1, B2, and C1 together. That’s 7 races. The winner and second-place finisher of this race are the second and third-fastest overall.

I don’t see a good way to do it in less than ten races.

I like a lot of @flutherother‘s setup, except for the multiple races for the top finishers in the first heat. That seems to militate against doing well in those heats. So I’d give the top finishers in the A-E heats a bye to their semi-finals run.

Here’s my whole solution:
5 heats, A – E: Keep A1 – 3, B1 – 3, C1 – 3, D1— 3, E1 – 3. The “1” horses get a bye to their semi-finals race.

Two more heats (F & G) with A2, A3, B2, B3, C2 and C3, D2, D3, E2, E3

Keep the top two finishers in each of these heats: F1 – F2, G1 – G2.

Run a semi-finals heat H consisting of A1, B1, C1, D1, E1. Keep the three fastest finishers.
Run a semi-finals heat J consisting of F1, F2, G1, G2. Keep the fastest two finishers.

Run a final race consisting of H1, H2, H3, J1, J2. First three over the line are the ones you want.

9 qualifying heats, with a single finals race.

@9doomedtodie‘s and @flutherother‘s solution is correct, with 7 heats.
I’m going to copy some of WasCy’s notation.

Run five heats, A – E.
Keep the top three horse from each race.

Race A1, B1, C1, D1, and E1, the winners from each heat. For simplicity, let A1 place 1st, B1 place 2nd, C1 place 3rd, etc.

Eliminate any horse that is slower than three horses. This means that all D and E horses can be eliminated, because even the fastest D and E horse still loses to at least 3 horses. C2 and C3 can be eliminated because C2 is slower than C1, which is slower than B1 and A1. B3 can be eliminated by the same logic.

This leaves A1–3, B1–2 and C1. We know that A1 is the fastest, so we no longer need to race him.

The five horses A2–3, B1–2, and C1 can be raced in one final race, from which we keep the top two horses, which are 2nd and 3rd fastest.

We have 5+1+1 = 7 races to find the top three horses.

As a person who has raced horses, I have a more logical but less elegant answer, run 5 races of 5 horses and clock them, then compare the times with each other

@flutherother yeah but if you race them in the same day that particular 4th horse can have a bigger stamina then the first 3 he raced with and if he competes with the rest 20 and races against first 3 only in the last round he can win .

So where the logic here ? He looses one round so he’s slower ?

To actually know which horses are the fastest you need to make all horses compete with each other several times and in different days or else I could just say ” it was just one of those days when he didn’t compete at his full potential ” ^^

@DrBill yeah but as you know if you put the same 5 horses compete at race track several times in a row or in different days you know it’s less likely that the same 3 horses will be in the first 3 positions .

Remember anyone can get it at any time .

@Hibernate

Using your logic, the top three can never be determined, because horses run different in different conditions

I know that’s why I put it this way . The puzzle doesn’t give a certain situation so we could interpret it as we see fit . [ ok I know I shouldn’t do this but I couldn’t help here ]

In any case I’ll curious to see how others try to get the answer [ besides the one that were already told ] .

This is a math puzzle and you are expected to make some simplifying assumptions to be able to apply mathematical reasoning. Otherwise a solution is not possible.

I am pretty sure that 7 is the best that can be done. It takes at least 6 races to determine the fastest horse, since each race eliminates exactly 4 possibilities for fastest horse. I don’t know of a similar simple way of determining how many races to determine second and third, but one extra seems to me as good as you can do.