(1) There are 25 horses. How will you identify the fastest three horses given that in a particular race, a maximum of five horses are allowed. A minimum of how many such races are required to get the result?
(2) A person has two numbers, he tells the sum of those numbers to person S and product of those numbers to P.
The following is a conversation between S and P:
S: I don’t know what the numbers are.
P: I don’t know what the numbers are either.
S: I know what the numbers are.
P: I know what the numbers are.
Assuming S and P are wise and good in Mathematics, what are those two numbers?
Note: Numbers are greater than 0.
(1) We can first analyse five races by taking five-on-five horses and get the top horse in each race.
Let’s say it looks like this:
O1, O2, O3, O4, O5
T1, T2, T3, T4, T5
TH1, TH2, TH3, TH4, TH5
F1, F2, F3, F4, F5
FV1, FV2, FV3, FV4, FV5
Here O1 > O2 > O3 > O4 > O5, similarly for others.
Now let’s take the fastest horse in each race and then have one race between them. So, have a race between O1, T1, TH1, F1 and FV1.
Let’s say O1, TH1 and F1 come in top three (and O1 > TH1 > F1), then there are no chances for horses slower than T1 and FV1 to be among the top three and we can also safely conclude that O1 is the fastest horse. Also, F2-F5 horses are not among the top three.
Now we need to find the second and third fastest horses. They can be either O2, O3, TH1, TH2 or F1. So we can have one more race among them to determine the second and third position. Thus, a minimum seven races are needed to find the top three horses from 25 horses.
(2) Numbers are 2 and 2.
S would have got 4, which means 1+3 or 2+2. That is why he was not sure of the numbers.
P would have also got 4, possibilities 1*4 and 2*2.
Now, had the numbers been 1 and 3, P would have got 3 and he would have been sure of the numbers, but that was not the case. So S became sure that the numbers are 2 and 2.
Later, P comes to know that the numbers cannot be 1 and 4, because there are two possibilities of getting the sum as 5, 1+4, 2+3, and in both these cases S cannot guess the number depending on P’s earlier answer, as for both products 4 and 6 there are more than one possibilities. Thus, P knows that the numbers are 2 and 2.