bhramarraj

1. Yes. 2. 12 times. as clock always show 12:00 whether it is midnight or is noon.

LED's are just there just to confuse the reader. And at 00.00 it will sound 12 times.

Suman has a very lovely battery driven wall clock in her bedroom. Every hour a cuckoo comes out of the clock and sounded as many times as is the time at that hour in a very sweet voice. Cuckoo sounds every 15 minutes also but for once only. Five LED's of different colors are provided in a circle, at the middle of the clock. LED's glow whenever cuckoo sounds, making the clock looks very beautiful. It was a daily routine of Suman to go to sleep at around 10:30 PM. On a hot night, when she was in deep sleep, she woke up suddenly feeling hot. She wondered what happened to A.C. In the dark she tried to switch on the light, and then she came to know that power supply was not available. Suddenly she heard cuckoo sounding once, LED's also glowed for once and went off. She could not see what time it was and wondered what time it was. She could not sleep due to heat, and had to wait for power supply to restore. After some time again the clock sounded once, LED's glowing for once, leaving Suman wondering how many times she has to listen the clock to know the exact time without seeing the clock.....? Could anybody guess it for Suman....?

Really brilliant thinking, though too long.

In the solution I have given solution to reach the top of both the poles. To reach top of only one other pole, you may apply the trick only for other pole. Is this the right solution..... or there is some other trick also...?

I am still confused, although I agree to your version that three cars can move on the 8 shaped track with the speeds in ratio 1:2:3, when it is allowed to cross other cars at X, irrespective of their directions at X. But how will it be possible to insert fourth car on the track...? The speed should fall between the ratio 1:2:3. But fourth car should also reach X when slowest car reaches X, and that too with the speed lesser than the second faster or third fastest car. Then what will happen...? I think the fastest car is definitely going to overtake fourth car, in between the track at a place other than X.

you are right. I missed this condition. Thank you.

Let the three sages be A, B, & C. 'A' sees the two blue hats on the heads of B & C. He waits and thinks, "What B & C are seeing..... if I have white hat, both will see one blue hat on each others head and one white hat on my head. Then, either will wait for answer from the other, both assuming the possibility of a white hat on their heads. But only I know that both of them have blue hats, so nobody will answer. Therefore, after hearing no answer from one another, B & C - both will be able to know that they did not have white hat, and quicker one will be able to tell his hat's color is blue.' But what if I had a blue hat...? In that case they will be thinking in the similar way as I am thinking, waiting response from either me or the other one. " After waiting for a calculated time, when 'A' hears no response from B & C, being the quickest, 'A' calls out first, "My hat's color is blue."

Solution is to by two mangoes so that when half is taken away and one is return quantity remains same as two at every floor. So you get exactly two mangoes at 7th floor for your wife.

I will put the solution this way: There may be three situations with different conditions. (1) Zab's face is black, while Rab's and Hab's faces are clean. Rab will see one black and one clean face, and he will be be laughing on Zab's black face. Hab will also see one black and one clean face, and he will also be laughing at Zab's black face. But Zab will see two clean and laughing faces, so he will not laugh and he will not be knowing why other two were laughing..! Rab and Hab will be able to conclude that since Zab is not laughing, so their own faces are clean. (2) Hab's & Zab's faces are black, and Rab's face is clean. So Rab is able to see two black faces. He will be laughing at the black faces of Hab & Zab, rightly thinking his own face was clean. Hab will see one black and one clean face, so he will laugh at the black face of Zab, thinking his own face was clean. Zab will also see one black and one clean face, so he will laugh at the black face of Hab, thinking his own face was clean. Now quicker Rab starts thinking "Hab & Zab are laughing at each other's black face and my face is clean?" Next fastest Hab thinks, "Why Zab is laughing while Rab's face is clean. It means my face is also black as of Zab." So realizing that his face is also black as of Zab, Hab will stop laughing. Then slowest Zab will think, "Hab was laughing while Rab's face is clean, it means my face is also black as of Hab. Now Hab has stopped laughing because he has realized now that his own face is also black." Thus Hab and Zab both realize that their faces are black. (3) Faces of all three--- Rab, Hab, and Zab--- are black. In the beginning, each of them thinks like situation (2) above, that faces of other two are black, and his own face is clean. So each of them laughs at black faces of other two. But quicker Rab starts thinking, "If my face is clean, and if Hab and Zab are laughing on each other's black faces, then why they are not able to conclude that the faces of both of them are black?" [Note that in the similar situations at Sr No. (2), Hab had been able to conclude that his own face was also black]. So Rab concludes that there could be only one reason, that he too have a black face so he stops laughing at other two. But next fastest Hab might misunderstand here. He may think, "Rab & Zab were laughing on each other's black faces and they are seeing my face is clean. Being quicker Rab has now concluded that his own face is also black and therefore has stopped laughing. So he continues laughing at black faces of Rab & Zab. Zab, the slowest, also thinks same way and comes to the wrong conclusion that his own face is clean. So he also contiues laughing at black faces of Rab & Hab. Now Hab quicker than Zab further anlyzes, "After seeing that Rab has stopped laughing, why did Zab not realize that his own face is also black..?" [see Sr Number (2)]. Then Hab realizes that the conditions are different from as at Sr no. (2) above, and only possibile condition left is that his own face is also black. Now slowest Zab is thinks "Rab stopped laughing realizing that his own face is also black like the black face of Hab, and now Hab also stopped laughing after realizing the fact that his face is also black. So my face is clean." So we see that the slowest thinker Zab may not be able to know about his own black face. PLEASE CORRECT ME IF WRONG.

"or him giving a box in return will hurt your teeth," this sentence makes the answer clear------Boxing..... where punch is indicated by 'Box'.

There is no possibility when only one person will be able to answer the color of his hat. Say A & B are on the right side of the wall and both of them know that one is wearing the heart shaped hat, and other one is wearing black or white hat. So if A sees heart shaped hat on B then he will not be able to determine color of his own hat, so he keeps silence. Thus, acknowledging silence of A, it becomes known to B that he is not wearing black or white hat, hence he is wearing heart shaped hat. So in this condition B can call his hat identity. Again, say A sees B wearing black or white hat, then he will be able to tell That his hat is heart shaped. Say C, D, & E are on the other side of wall. Say E sees two white or two black hats in front of him, then he will be able to call out his own hat's color. And if E sees one black & one white colored hats in front of him, then he will not be able to able to call out his hat's color, but at the same time acknowledging the silence of E, it becomes known to D that he himself and C are wearing hats of different colors. So D is able to call out his hat's color which must be black if white is seen on the head of C, and white if black is on the head of C. So we see that in either conditions at least two persons will be able to identify their hats.

Hope you dont mind me posting this