Brainteasers/Math problems - Page 9

Lets work it backwards, but to make sure that all conditions will be properly executed.
When there is 1 person - he gets all.

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When there are 2 people. 1st get 0, 2nd gets 100.
1st one votes - No, 2nd votes Yes.
1:1 votes - split accepted!

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When there are 3 people, 3rd person will offer - irrelevant.
1st knows he can't get anything, so he will for the vote to fail, in order to kill more pirates.
2nd knows he can get 100 and stay alive, so he will vote to fail, in order to satisfy last condition - kill more.
3rd person - what ever he offers doesn't matter.
1:2, split fails.

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When there are 4 people:
4th will propose: 1 - 0, 2-0, 3 - 0, 4-100

1st knows he can't get anything, so he will for the vote to fail, in order to kill more pirates.
2nd knows he can get 100 and stay alive, so he will vote to fail, in order to satisfy last condition - kill more.
3rd will try to stay alive, because if the vote fails, he will be killed. He will vote yes.
4th will stay alive and will try to make the most of the money. He will vote yes
2:2 split accepted.

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When there are 5 people:
5th will propose - irrelevant.

1st knows he can't get anything, so he will for the vote to fail, in order to kill more pirates.
2nd knows he can get 100 and stay alive, so he will vote to fail, in order to satisfy last condition - kill more.
3rd will try to stay alive, because if the vote fails, he will be killed. He will vote yes.
4th will try to kill more and he know he can stay alive. He will vote No.
5th will vote yes, to stay alive.
2:3 - split denied.

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When there are 6 people:
6th will propose

1st knows he can't get anything, so he will for the vote to fail, in order to kill more pirates.
2nd knows he can get 100 and stay alive, so he will vote to fail, in order to satisfy last condition - kill more.
3rd will can safely vote No, because he can safely kill more pirates.
4th will try to kill more - what he gets doesn't matter. He will vote No.
5th will vote yes, in oder to stay alive.
6th will try yo stay alive. He will vote Yes"
2:4- split denied.

At this point, there is no point to keep going onward.
The final answer is
there will be 4 pirates left, and 4th one (counted backwards) will get all the money.

Fill in the 5 liter, then spill it into the 3 liter. When the 3 liter is full, you have 2 liters left in the big one. Do it twice, and you got 4 liters.

Take 1 from the first bag, 2 from the second bag, 3 from the third, etc... Multiply the number of grams missing (the 55 (1+2+3+...+10=55) coins should weigh 55 grams if they were all real) by 10 and you have the number of the bag with fake coins.

Do the same thing with 1/2/4/8 coins from bags #1/2/3/4.

Flip them both. When 7 is empty, flip the 7. 7 minutes have gone by. When 11 is empty, 11 minutes have gone by, and 4 minutes since you flipped the 7. Flip the 7 again, and wait for it to be empty. 15 minutes have gone by.

It's easy to find that it's around 1h05. To find the exact time, lets call x the angle that the hands do with the 12. Hour and Minutes are on the x. "60+x minutes" has to be equal to "x hours".
"x hours" is x*12/360 hours, or x*12/6 minutes, or 2x minutes (for example, 90° = 180 minutes = 3 hours). "x minutes" is x*60/360 minutes or x/6 minutes (for example 90° = 15 minutes). So we need to have 60+x/6 = 2x, which gives 11x/6 = 60 or x = 360/11 degrees.
Both hands are on 360/11 degrees, which means it's 1:05:05

All 100 guys with blue eyes will leave on the 100th day.
Let's assume there's only one guy with blue eyes. He'll look around him and see no one has blue eyes, but he knows someone does, so he knows it has to be him, so he leaves on the first day.
Now, if there was 2 guys with blue eyes, each of them would think "This dude is the only one I can see who has blue eyes. If I don't have blue eyes, he doesn't see anyone with blue eyes, so he'll leave. Let's see if he's still here tomorrow..." The next day, the other dude's still here, so it means I too have blue eyes. The other guy thinks the exact same thing, and we both leave on the 2nd day.
[...]
Now, with 100 guys with blue eyes, each one of them will think "I see 99 guys with blue eyes. If I don't have blue eyes, they'll all leave on the 99th day. Let's see if that happens. On the 100th day, the other 99 guys are still here, which means I too have blue eyes. The other 99 guys think the same thing, we all figure out we have blue eyes, and all leave on the 100th day.

Split them in 3 piles of 3. Weigh one of the piles against the other. 2 possibilities :
- it's balanced : the lighter coin is in the 3rd pile
- it's not balanced : the lighter coin is on the lighter side
Either way, we now know in which pile of 3 the lighter coin is.
Weigh one coin of this pile against another of this pile.
- it's balanced : the 3rd coin is lighter.
- it's not balanced : we found the lighter coin.

Light one on one end, and the other on both hands. When the one you lighted on both end is done burning, 30 minutes have gone by. Light the other hand of the first one, when it's done burning, 15 more minutes have gone by, total 45 minutes.

Lets call A our prime number greater than 32.
R < 30. Lets suppose R is not a prime number.

The first vizir makes 2 steps forward. The second vizir comes close to the other. Now the third vizir has 2 options :
- if the first 2 have different colors, he goes between them
- if the first 2 have the same color hats, he goes to either end of this new line.
The 4th vizir does the same :
- if all 3 hats he sees in the new line are the same color, he goes to one end of the line
- if there are hats of both colors, he goes between the red and blue that are next to one another.
In the end, if they all do that, all reds will be on one side, all blues on the other. They will all announce their color starting from the most left, then the most right, then the second-most left, then the second most-right. Only one will be wrong (or maybe they'll all be right, if they're lucky).

Yes you should switch.
If you picked the door with the car first, switching means losing and not switching means winning.
If you picked the door with the goat first, switching means winning and not switching means losing.
But the second situation happens twice more often, which means you double your chance of winning by switching (2/3 vs 1/3).

Of course, it's exactly like writing in binary.

45? Each line can cross every other once, and I don't think there's a reason why some intersection points should be common, so for each line there is 9 intersection points. But you must not count twice the point of intersection of line A and B and line B and A, so half of 10*9. It seems too easy, maybe there's a catch.

http://xkcd.com/solution.html

Very interesting and much more simplistic than I thought it would be.

If there's you, B(lue)1, B(lue)2, B(lue)3. Now there is 2 cases, either you have blue eyes or you don't. If you don't have blue eyes the only people with blue eyes that B2 will see is B1 and B3. B2 knows that if he didn't have blue eyes B1 and B3 would have left on the second night. If B1 and B3 doesn't leave on the second night B2 knows that he must have blue eyes aswell. And if B1, B2 and B3 doesn't leave on the third night then you know that you must be blue eyed. Using this logic you can figure it out.

The key is not thinking that they see eachother, instead take it in steps thinking, "If I don't have blue eyes, B1 only sees 2 people with blue eyes" and then you put yourself in the perspective of B1 with the assumption that you do not have blue eyes which would lead to him thinking "If I don't have blue eyes, B2 only sees 1 person with blue eyes which would make him leave on the second night"

Number1 and Number2 forward - 2min
Number2 back - 4min
Number3 and Number4 forward - 14min
Number1 back - 15min
Number1 and Number2 forward - 17min

You will not be able to see the hat on your head.
Is the statement you should focus on, if you are smart enough you'd just take the hat off your head so you can see it.
These riddles usually go with something so obvious that you feel a little wetarded when you have to solve it.

Reminds me on "How fast can everyone touch the ball trick?" Our class of about 30 students all touched it in about 1 seconds flat after we figured it out, first we were passing the ball to each other got that down to about 20 something seconds, then we tried getting closer, got it down more, then we tried just handing it off in a circle got it down faster. Then we all put a finger on the ball and let go at the same time.

It just shows there is more than one solution, but which is best? In other words a lot of the solutions weren't wrong, and a lot weren't right, the ones where you have to sacrifice someone is just silly =P Nobody needed to die! just take the hat off and put it on your hand and say the color

The 2$ in the boy's pocket was paid by the men, it is already in the 3*9$ they paid. The missing 3$ was given back to the men, 3*9$+3$=30$. Deceptive puzzle, but can't be misunderstood when read online.

After a grueling process of keeping track of every possible offer each pirate member can have for every split, I've finally come to a conclusion. So working backwards, we start with the case with 2 pirates left. The one doing the splitting will be labeled as pirate#2, with the one lower in rank being pirate#1 (so the rest will be labeled in accordance to their ranks with the captain being pirate#10. As it has been discussed before, #2 has the power to split everything to himself, since his vote is enough as the majority. so:

Split1: #2-100, #1-0

From here a lot of reasoning will be cumulative and dependent upon the possible offers of previous splits. So now when #3 is doing the splitting, he can easily persuade #1 to agree with him with one coin, since it's better than getting 0 with #2 splitting. #2 will be inclined to always disagree until it is his turn to split, but his vote does not matter. #3 and #1 are enough as the majority. So:

Split2: #3-99, #2-0, #1-1

So #4 has to try and persuade at least 1 person to his side yet again like with #3. #2 and #3 are going to disagree no matter what since they stand to benefit much greater when it is their turn. So similarly with #3, he can just offer #1 a bit more than what #3 will have to offer.

Split3: #4-98, #3-0, #2-0, #1-2

Now things change up. #5 is in a bad position as #1 isn't enough for majority. In fact, #4, #3, and #2 will benefit tremendously from their splits only, so they'll never agree. so #5 is destined to die.

Split4: #5 will die

Things change greatly again. #6 is in a much better position than #5, since he can offer #5 0 as #5 will still have his life, and #1 a bit more than what #3 would have offered for majority!

Split5: #6-97, #5-0, #4-0, #3-0, #2-0, #1-3

Now like #5, #7 is also screwed since he can not ever get #6, #4, #3, or #2 to agree. He'll never reach majority and die.

Split6: #7-dies.

Do you see the pattern now? The odd ranks greater than 3 can not pull majority no matter what. They will die when they try to split. The even ranks can abuse this, offer minimum to those destined to die and #1 a bit more than what they would get so far. So moving along, #5 accepts 1 as it's now the best offer he can get after just his life in the previous offer. #7's new best offer is simply his life. This pattern will repeat itself until it's the captain doing the splitting.

Split7: #8-95, #7-0, #6-0, #5-1, #4-0, #3-0, #2-0, #1-4

Split8: #9 will die

Split9: #10-92, #9-0, #8-0, #7-1, #6-0, #5-2, #4-0, #3-0, #2-0, #1-5

The end result is one cunning captain with most of the treasure for himself, and the majority formed by #9, #7, #5, and #1 sucking it up and agreeing since there's absolutely no better possible outcome for them unless they agree.

One important possible flaw: I'm assuming the pirates will only accept the offer if it's GREATER than any possible future offer, not equal to. If the pirates are willing to agree even if future offers are the exact same, then the captain would just end up with all 100 and still get his majority to agree.

U could ask do u think lying to someone is ok?
If u pick the younger one she will tell u no(lie)
If u pick the older one she will tell u no
If u pick the middle one she will answer yes because she lies sometimes and tell the truth other times

U then know if she answers yes that she is the middle sister and if she answers no u marrie her

U start by balancing 4 to 4 if it is equal u have 4 left and in those 4 is the coin u are looking for. U then balance 2 of those coins with 2 good ones, if it is not balanced u know that in those 2 is the bad one, then u take one of them and balanced with a good one. if it is equal u know the last one is the bad coin and if it is unbalanced u know that one is the bad coin. If on the second balance the 2 pair of coins are balanced u know the last 2 are the bad ones and u repeat a few steps up. Now comes the hard part if u have it unbalanced at the start 4 to 4