Nine IITians were sitting in a classroom. Their prof wanted to test them. One day he told all of his 9 students that he has 9 hats, colored either in red or black. He also added that he has at least one hat with red color and the no. of black hats is greater than the no. of red hats. He keeps those hats on their heads and ask them tell me how many red and black hats I have? Obviously students can not talk to each other or no written communication, or looking into each other eyes; no such stupid options.

Prof goes out and comes back after 20 minutes but nobody was able to answer the question. So he gave them 10 more minuets but the result was the same. So he decides to give them final 5 minutes. When he comes everybody was able to answer him correctly.

Prof goes out and comes back after 20 minutes but nobody was able to answer the question. So he gave them 10 more minuets but the result was the same. So he decides to give them final 5 minutes. When he comes everybody was able to answer him correctly.

So what is the answer? and why?

See Solution : Hats and IIT studentsAfter first interval of 20 minutes :

Lets assume that their is 1 hat of red color and 8 hats of black color. The student with red hat on his head can see all 8 black hats, so he knows that he must be wearing a red hat.

Now we know that after first interval nobody was able to answer the prof that means our assumption is wrong.

So there can not be 1 red and 8 black hats.

After second interval of 10 minutes :

Assume that their are 2 hats of red color and 7 hats of black color. The students with red hat on their head can see all 7 black hats and 1 red hat, so they know that they must be wearing a red hat.

Now we know that after second interval nobody was able to answer the prof that means our assumption is again wrong.

So there can not be 2 red and 7 black hats.

After third interval of final 5 minutes :

Now assume that their is 3 hats of red color and 6 hats of black color. The students with red hat on their head can see all 6 black hats and 2 red hats, so they know that they must be wearing a red hat.

Now we know that this time everybody was able to answer the prof that means our assumption is right.

So there are 3 red hats and 6 black hats.

Now as everybody gave the answer so there can be a doubt that only those 3 students know about it how everybody came to know ?

Then here is what i think, the professor gave them FINAL 5 minutes to answer, so other guys will think that the professor expects the answer after 3rd interval (according to prof it must be solved after 3 intervals), so this is the clue for others.

Facebook Comments

DeePAK says

There are two easy way in which professor could put the hat and 1 semi complicated and one complicated way.

1R – 8B obviously easiest combo red guy would know.

4R – 5B – again all black guys would know.

*****if they are not out that means first interval gives the clue.

Now 2R – 7B , each of the red. Guy know that if I have black the other red guy would have shouted. So I must be red and they’ll shout.

So the only option left is last one. no clue will be the biggest clue in this case.

Mantis says

What a bullshit answer to this question!

The correct answer for why they needed 35 minutes to answer this question is that maybe because they got into the IITs in their third attempt.

From the question, we know the following:

1) We have 9 Hats.

2) |R| >= 1

3) |R| > |B| (strictly)

The following cases are possible for (R,B) : {(1,8), (2,7), (3,6), (4,5)}

Cases (1,8) & (4,5) should be trivial – For (1,8) the Red Guy can count the number of Black hats (which is 8) and knows there is a minimum of 1 Red Hat. So he answers first.

For (4,5) – The Black guy answers first as he will see equal number of Red & Black Hats and knows |R| > |B|.

Let us conside Case (3,6):

A Red Guy will see 2R + 6B hats. He won’t be able to decide his hat, as it could be Red or Black, meeting the specifications above.

Let us exclude two black guys of the lot (B1 and B2). Both see 3R + 4B hats.

Of these two, our first Black guy B1, will see 3R + 5B hats. He will think his hat could be Red or Black, as both confirm to the requirements above. The second black guy B2, sees that the first black guy is not able to answer. B2 will think if his hat were Red, B1 should have seen 4R + 4B hats, and should have answered. But he didn’t. So B2 should realise after a while since B1 is not answering his hat has to be Black. And hence answers.

The only case where no-one would be able to answer will be when there are 2R and 7B Hats.

Rahul says

Bravo..!

Binayak says

Case 1. 1 Red Hat 8 Black Hat

The guy with Red Hat can answer it in second.

Case 2. 4 Red Hat 5 Black Hat

Students with Black Hat can answer in second.

Case 3. 2 Red Hat 7 Black Hat

Students with Red Hat can answer in minute.

Students with Red Hat will wait for second to see whether the other

student with Red Hat is coming with the answer (1,8) or not if he does

not then he must be seeing a Red Hat.

Case 4. 3 Red Hat 6 Black Hat

Students with Red Hat can answer in minute

Students with Red Hat will wait for minute to see whether the other

students with Red Hat is coming with the answer (2,7)(as above) or not

if he does not then they must be seeing two Red Hat. Wich mean there are

three Red Hat.

Honest One says

Not too obvious..

If we are taking the approach of elimination then pretty obviously (8B,1R) and (5B,4R) are out of equation since they are too obvious. Now lets take the case of(7B,2R). If you think like one of the red(hat) guy, he will be looking at the combination of(7,1) and since the other red guy also didn’t answer the question in first go that means he was not able to make decision and hence in second go it has to be (7,2) (Same thinking could be applied as a perspective of a black(hat) guy). For me 6,3 and 7,2 both are ambiguous.