Two bodies have I Though both joined in one. The more still I stand The quicker I run. What am I?
Answer: I am an hourglass.
Solution:
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3 persons crossed the bridge, who are these people riddle with answer
Answer: A pregnant women with one baby on her arms
Solution:
Riddle with answer: The first person saw the bridge step on it and crossed, the second person saw the bridge did not step on it but crossed, the third person did not see the bridge did not step on it but crossed. how it is possible and Who are these people?
we have explained the answer to Who are these people riddle, after watching the video you be shocked with the correct answer. This person crossed a bridge puzzle little tricky so you need to think in out of the box.
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Correct answer: A pregnant women with one baby on her arms
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Riddle with answer: The first person saw the bridge step on it and crossed, the second person saw the bridge did not step on it but crossed, the third person did not see the bridge did not step on it but crossed. how it is possible and Who are these people?
we have explained the answer to Who are these people riddle, after watching the video you be shocked with the correct answer. This person crossed a bridge puzzle little tricky so you need to think in out of the box.
Show answer
Correct answer: A pregnant women with one baby on her arms
Knowledge increases by sharing 🙂
Before he turned physics upside down, a young Albert Einstein supposedly showed off his genius by devising a complex riddle involving a stolen exotic fish and a long list of suspects. Can you resist tackling a brain teaser written by one of the smartest people in history? Dan Van der Vieren shows how.
Answer: The key is that the person at the back of the line who can see everyone else's hats can use the words "black" or "white" to communicate some coded information. So what meaning can be assigned to those words that will allow everyone else to deduce their hat colors? It can't be the total number of black or white hats. There are more than two possible values, but what does have two possible values is that number's parity, that is whether it's odd or even. So the solution is to agree that whoever goes first will, for example, say "black" if he sees an odd number of black hats and "white" if he sees an even number of black hats. Let's see how it would play out if the hats were distributed like this. The tallest captive sees three black hats in front of him, so he says "black," telling everyone else he sees an odd number of black hats. He gets his own hat color wrong, but that's okay since you're collectively allowed to have one wrong answer. Prisoner two also sees an odd number of black hats, so she knows hers is white, and answers correctly. Prisoner three sees an even number of black hats, so he knows that his must be one of the black hats the first two prisoners saw. Prisoner four hears that and knows that she should be looking for an even number of black hats since one was behind her. But she only sees one, so she deduces that her hat is also black. Prisoners five through nine are each looking for an odd number of black hats, which they see, so they figure out that their hats are white. Now it all comes down to you at the front of the line. If the ninth prisoner saw an odd number of black hats, that can only mean one thing. You'll find that this strategy works for any possible arrangement of the hats. The first prisoner has a 50% chance of giving a wrong answer about his own hat, but the parity information he conveys allows everyone else to guess theirs with absolute certainty. Each begins by expecting to see an odd or even number of hats of the specified color. If what they count doesn't match, that means their own hat is that color. And every time this happens, the next person in line will switch the parity they expect to see.
Solution:
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The key is that the person at the back of the line who can see everyone else’s hats can use the words “black” or “white” to communicate some coded information. So what meaning can be assigned to those words that will allow everyone else to deduce their hat colors? It can’t be the total number of black or white hats. There are more than two possible values, but what does have two possible values is that number’s parity, that is whether it’s odd or even. So the solution is to agree that whoever goes first will, for example, say “black” if he sees an odd number of black hats and “white” if he sees an even number of black hats. Let’s see how it would play out if the hats were distributed like this. The tallest captive sees three black hats in front of him, so he says “black,” telling everyone else he sees an odd number of black hats. He gets his own hat color wrong, but that’s okay since you’re collectively allowed to have one wrong answer. Prisoner two also sees an odd number of black hats, so she knows hers is white, and answers correctly. Prisoner three sees an even number of black hats, so he knows that his must be one of the black hats the first two prisoners saw. Prisoner four hears that and knows that she should be looking for an even number of black hats since one was behind her. But she only sees one, so she deduces that her hat is also black. Prisoners five through nine are each looking for an odd number of black hats, which they see, so they figure out that their hats are white. Now it all comes down to you at the front of the line. If the ninth prisoner saw an odd number of black hats, that can only mean one thing. You’ll find that this strategy works for any possible arrangement of the hats. The first prisoner has a 50% chance of giving a wrong answer about his own hat, but the parity information he conveys allows everyone else to guess theirs with absolute certainty. Each begins by expecting to see an odd or even number of hats of the specified color. If what they count doesn’t match, that means their own hat is that color. And every time this happens, the next person in line will switch the parity they expect to see.
Show Answer
The key is that the person at the back of the line who can see everyone else’s hats can use the words “black” or “white” to communicate some coded information. So what meaning can be assigned to those words that will allow everyone else to deduce their hat colors? It can’t be the total number of black or white hats. There are more than two possible values, but what does have two possible values is that number’s parity, that is whether it’s odd or even. So the solution is to agree that whoever goes first will, for example, say “black” if he sees an odd number of black hats and “white” if he sees an even number of black hats. Let’s see how it would play out if the hats were distributed like this. The tallest captive sees three black hats in front of him, so he says “black,” telling everyone else he sees an odd number of black hats. He gets his own hat color wrong, but that’s okay since you’re collectively allowed to have one wrong answer. Prisoner two also sees an odd number of black hats, so she knows hers is white, and answers correctly. Prisoner three sees an even number of black hats, so he knows that his must be one of the black hats the first two prisoners saw. Prisoner four hears that and knows that she should be looking for an even number of black hats since one was behind her. But she only sees one, so she deduces that her hat is also black. Prisoners five through nine are each looking for an odd number of black hats, which they see, so they figure out that their hats are white. Now it all comes down to you at the front of the line. If the ninth prisoner saw an odd number of black hats, that can only mean one thing. You’ll find that this strategy works for any possible arrangement of the hats. The first prisoner has a 50% chance of giving a wrong answer about his own hat, but the parity information he conveys allows everyone else to guess theirs with absolute certainty. Each begins by expecting to see an odd or even number of hats of the specified color. If what they count doesn’t match, that means their own hat is that color. And every time this happens, the next person in line will switch the parity they expect to see.
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