This is a classic puzzle which can look simple.

Given 25 horses, find the best 3 horses with minimum number of races. Each race can have only 5 horses. You don't have a timer.

Now the challange is how we can do it in 7 races.

Solution

We will have 5 races with all 25 horses

Let the results be

a1,a2,a3,a4,a5

b1,b2,b3,b4,b5

c1,c2,c3,c4,c5

d1,d2,d3,d4,d5

e1,e2,e3,e4,e5

Where a1 faster than a2 , a2 faster than a3 etc and

We need to consider only the following set of horses

a1,a2,a3,

b1,b2,b3,

c1,c2,c3,

d1,d2,d3,

e1,e2,e3,

Race 6

We race a1,b1,c1,d1 abd e1

Let speed(a1)>speed(b1)>speed(c1)>speed(d1)>speed(e1)

We get a1 as the fastest horse

We can ignore d1,d2,d3,e1,e2 and e3

a2,a3,

b1,b2,b3,

c1,c2,c3,

Race 7

Race a2,a3,b1,b2 and c1

The first and second will be second and third of the whole set

## Thursday, November 3, 2011

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## 6 comments:

"The second and third will be second and third of the whole set"

shouldn't it be "The FIRST and SECOND will be second and third of the whole set" ?

Thanks Shamsi,

I have corrected it.

RACE#1 - 1 2 3 4 5

RACE#2 - R1 6 7 8 9

RACE#3 - R2 R2 10 11 12

RACE#7 - R3 R3 R6 R6 15RACE#6 - R5 R5 14 15 javascript:void(0)16

RACE#5 - R4 17 18 19 20

RACE#4 - 21 22 23 24 25

The R#(Race#) represent the winner of the previous race.

in race 1 - 3 we get the 2 best horse

in race 3 - 6 we also get the 2 best horse

so in race 7, 2 best horse in round 1-3 + 2 best horse in 4-6 + the one that did not yet compete

Why didn't you consider b3 too as the b3 < c1 relationship is not established yet??

@Akshay,

Why didn't you consider b3 too as the b3 < c1 relationship is not established yet??

Because b3 is slower than a1, b1 and b2 so it can never be in the top 3.

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