Blind Man

Keshav and Venkatesh, both blind from birth were having a casual talk. Keshav asked Venkatesh “Venkat, according to you what is color?” Venkatesh replied “There are three types of questions:
i) The questions for which the answers are known.
ii) The questions for which the answers are unknown.
iii) The questions for which the answers are unknowable.
Your question is one of the third type of questions. Try another.”
Keshav started asking another one “A blind beggar was begging in a village. One rich person came to him and threw 1000 coins in front of him. He told the beggar that out of those 1000 coins 200 coins are facing up. He asked the beggar to make two groups of these coins such that each group would have same number of coins facing up. If the beggar does so he will get all the coins. Now how should the beggar proceed?”


Keshav’s question is for you too. How should the beggar proceed?


Hint 1

It is not necessary that each of the two groups should have 100 coins facing up.


Hint 2

Think of flipping coins.


Answer

The beggar has to divide the 1000 coins into two arbitrary groups of 200 coins and 800 coins. Now he has to flip all 200 coins in the 200 coins’ group. Problem solved.


Solution

The beggar has to divide the 1000 coins into two arbitrary groups of 200 coins and 800 coins. Now he has to flip all 200 coins in the 200 coins’ group. Problem solved.



Generally people think that the beggar has to divide the group into two groups such that each group has to contain 100 heads. The problem with this approach is that it is impossible to know which coins are facing up. So this approach will not lead us to the solution.

If we think that it is impossible to continue from a point where we have stuck, then we have to try another approach. Another common approach is this: Consider that the beggar divides the 1000 coins into two random groups, one having x coins and other having (1000 - x) coins. We do not know how many heads are there in x coins’ group. Let it be y. Then we have this information.

	Group 1 with x coins	Group 2 with (1000-x) coins
Number of heads	y	(200 – y)
Number of tails	(x - y)	(800 - (x - y))

Now we know that it is impossible that the number of heads in both the groups is equal in all cases as we have reasoned it before. At this point of the solution we should get a sparkling idea that if all coins of a group is flipped, then we may get close to the problem’s destiny. Let us try. Consider the two cases:

Case I)
We flip all coins of Group 1, and then by equating the number of heads, we have
(x - y) = (200 - y) which implies that x = 200.

Case II)
We flip all coins of Group 2, and then by equating the number of heads, we have
y = (800 - (x - y)) which implies that x = 800.

In simple words, in both cases, we have to flip all the coins of the group with 200 coins. At this point we will have equal number of heads in both the groups, and they may not be equal to 100.

Generalizing the problem, if we have m coins in which n coins are facing up, where m and n are even, then divide the m coins into two groups of n and (m - n) coins. Now flip all the coins in the n coins’ group. Try it out.