## 25 Horses Problem-Solution

**Solution :**

7 races..

*Explanation:*

Divide the set of 25 horses into 5 non-overlapping sets of 5

horses each. Have a race each for all the horses in each

set. This makes it a total of 5 races, one for each set.

Now, have a race for the winners of each of the previous 5

races. This makes it a total of 6 races.

Observe the position of each horse in the 6th race and

correspondingly number the sets. i.e. the set of the winner

of 6th race will be said to be set no. 1 while the set of

the loser of the 6th race will be said to be set no. 5.

Now, possible candidates for the first three positions

exclude the followings:

1. Any horse from set 4 or set 5.

2. Any horse except the winner from set 3,.

3. Any horse except the winner and the 2nd position from set 2.

4. Any horse except the winner, 2nd position and 3rd

position from set 1.

So now we have 6 candidates for top 3 positions. However, we

know that the winner of set 1 is the fastest horse in the

whole group of 25 sets.

So now we have 5 candidates for the second and third

position. What better way to find out who’s who than to have

a race of these 5 horses. Race them and this will solve our

problem in just 7 races.

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Can you explain why you can take the following steps:

2. Any horse except the winner from set 3,.

3. Any horse except the winner and the 2nd position from set 2.

4. Any horse except the winner, 2nd position and 3rd

position from set 1.

How do you know, for example, that the second position from set two is faster than the second position from set three? Couldn’t the first position from set two be faster than all the positions in set three but any position in set three still be faster than the second position in set two?

Andrew,

I think you are under confusion. Let me reframe the points #2,#3 and #4.

I hope you got the idea, why we excluded the set #4 and set #5. The reason is, the best horse in the set itself is not as good enough to reach top 3. So none of the horses in the sets #4 and #5 could reach. So we ignored those ten.

Now coming to the point #2, explanation is very similar to previous one. If the best horse of the set #3 itself has got 3rd place only. So the best position that 2nd horse of set#3 can reach is 4th place. But we are bothered only about top 3. So we can ignore any other horse in set #3.

For point #3, In set #2, best case, 2nd horse can be the 3rd best. So we might have to consider that.

For point #4, In set #1, best case, 2nd and 3rd horse can reach in 2nd and 3rd place respectively, so we take them both into picture.

Hope you are clear now. If not respond me I ll try to explain it better :)

me too got clear….thanks…nice….