Brainteasers/Math problems - Page 8

Let's number the pirates A,B,C,D,E,F,G,H,I,J with A being the captain and so on.

It is ever gets to I to make the split, he will die automatically since J will disagree. So I will automatically agree with H, and so H can do whatever he wants, i.e. take all the money.

Thus, it is in I and J's advantage to agree with G as long as they get more than 0 (otherwise they might as well watch G get killed). Thus if it ever gets to G to make the split, he will be able to get away with this distribution (from G to J): 98-0-1-1. (H will get nothing because whatever offer G makes to him, he will disagree since it is in his advantage (H's) to disagree and become the one to make the split)

F can prevent this by offering this distribution (from F to J): 97-0-1-0-2 OR 97-0-1-2-0. From all steps here onwards, there is more than 1 choice, so the final distribution is not a unique solution. I will pick the former (97-0-1-0-2) in this step and pick an arbitrary choice for the following steps, which are as follows:
A--B--C--D--E--F--G--H--I--J
----------------96--0--1--2--1--0 All the 0's disagree while all the nonzeros agree, since they will get more than they will if it goes to the next split.

------------96--0--1--2--0--0--1

--------95--0--1--2--0--1--1--0

----95--0--1--2--0--1--0--0--1

94--0--1--2--0--1--0--1--1--0.

So the captain can always get away with $94. The captain has many choices as to how to split the rest which I will not bother to list.

Convert any number into base 2, and you'll get the powers of 2 you need to summate. Since each base 2 number corresponds with a single base 10 number, each expression is unique

IE:
7 = 111 base 2 = 4 + 2 + 1
14 = 1110 base 2 = 8 + 4 + 2

15 integers in any collection of 100 integers must have of the same equivalence class mod 7.

Assume there's only 14 of each class. 7 * 14 = 98 numbers total. Thus, we must add in two more numbers. Since each number must fall into one of the 7 equivalence classes, at least one class will have at least 15 numbers. Thus, we can pick those 15 numbers, and the terms of the problem will be satisfied

QED

Fill the 5 liter jug. Empty it into the 3 liter one. Empty the 3 liter jug in the pool. Empty the remainder of the 5 liter jug in the 3 liter jug. Fill the 5 liter jug. Fill the 3 liter jug with the 5 liter jug until it is full (that is, drop 1 liter from the 5 liter jug into the 3 liter jug). You have 4 liters in the 5 liter jug.

Place one bag on the scale, then another one.
If the scales are even, you don't have the 9 grams bag. If they are not, you found it (ligher one).
Rince and repeat, adding two additional bags each time'till you come to the point where you have the same number of bags on each side but the scales are uneven.
The moment you add a pair of bags and the scales are uneven, the last one you added on the ligther side is the 9 grams bag.

Start both hourglasses at the same time. Start boiling the egg when the 7 minutes hourglass ends. When the 11 minutes hourglass ends, reset it and wait until the end. Take the egg off the fire when this hourglass ends.

The hour hand goes 12 times slower than the minute hand. Therefore the next time they cross wil be between 1 and 2.
Let x be the minute differential to 1 when the hands cross. Let's consider the circle has a size of "1". 1/12 + x/(12*60) = x/60 <=> x=60/11=~5.45.
The hands will cross at 5.45 past one.

Woah

Yes,
Place six coins on the scales
If it is even, then put two of the remaining coins on the scales. If it is even, the counterfeit one is the one that is left. If it is not, it's the coin on the lighter side.
If is not, then take the three coins from the lighter side of the scale and do what we would have done before with the three remaining coins.

Yes,
Let p be a prime number greater than 32.
There exists a unique (q,r) couple of integers that are so that p =30q+r

Suppose that r is not a prime number.
Then there exists (a,b) couple of integers that are so that r=ab and a>1, b>1. What's more, a and b are odd, because p is odd (because any even number greater than 2 is not prime)

So, granted that ab is stricly smaller than 30, here are the possible situations :
a=3,b=w/e, then p=3(10q+b) and p is not prime. That is absurd.
a=5, b=w/e, then p=5(6q+b) and p is not prime. That is absurd.
a=7, then b is smaller than 5 (because 5*7=35), so b=3 because b is odd and b is not equal to 1 and we have the same result.

Therefore, it is absurd that r is not a prime number.
Hence, r is a prime number.

I don't think so. A 13-gon has a total of 13 corners. If you use parallelograms to tile it, you will end up having an even number of corners.

Yes. The probability that you picked the correct door is 1/3. The probability that the correct door is the other one knowing that the 3rd door contained a goat is 1/2

Let's prove the following property by induction: P(n):"For any integer p between 2^n and 2^(n+1), there exists an unique sum of distinct powers of 2 so that p is equal to that sum"

For n=0
1=2^0
2=2^1
Hence, P(1) is true.

Let n be an integer. Let's suppose that P(0),...,P(n) are true and let's prove P(n+1).
Let p be an integer between 2^n and 2^(n+1).

If p=2^n the problem is solved.

Otherwise, p-2^n is an integer between 1 and 2^(n)(2-1)=2^n.
Hence, there exists an unique expression as the one we are looking for, by induction.

Hence the property for any n integer.

I'm still very unclear about the cases >3
(does not really matter how many other colours are around.)

I'm going to use I in place of the observer. (blue eyes)

Case 1---> 1 blue eyes; nobody else has blue eyes. I must have blue eyes and leave.
Case 2 --->2 blue eyes; I see the guy with blue eyes did not leave, I must have blue eyes as well.
Case 3 --->3 blue eyes; I see two other people with blue eyes, they must also see at least 1 set of blue eyes. I can see two sets of blue eyes, eliminating the possibility that they must see MORE than 1 set of blue eyes.

Why do I assume that those two people can see more than 1 set?

Thats the part I don't understand really I guess...

There are 2 pirates left (8 die). The ninth pirate will get all the gold.

Basically to satisfy all conditions it will go like this:

10= 9x11 + 1x1
9= 8x12 + 1x4
8= 7x14 + 1x3
7= 6x16 + 1x4
6= 5x20 + 1x0
5= 4x25 + 1x0
4= 3x33 + 1x1
3= 2x25 + 1x0
2= 1x100 + 1x0

To satisfy condition #1, the captain will offer whatever he can to say alive (At the maximum he can offer its 11 pieces to everybody. Anything else is irrelevant). The rest of the crew minus the next in command can safely vote to fail, since condition #1 is true and they try to achieve condition #2 - get more money and condition #3 - kill more pirates.

The cycle repeats itself until there are only 2 people left. At this points all conditions are true.

Half of the crew will agree this is the split: 9th pirate will propose that 9th gets 100, 10th get 0 and the split is approved. The last person can no break this as the split simple gets approved with 50 per cent majority.

The treasurer realized he would never be able to make a stronger poison - That the pharmacist's poison would always be stronger then his own. So he instead makesa solution that has no poison in it at all, then before the contest starts drinks a poison weaker then the pharmacist's would be. This way he will survive having his weaker poison neutralized by the pharmacist's stronger poison and the pharmacist would die from his own poison. The pharmacist realizes this and instead of making a really strong poison makes a solution of no poison at all as well. So that when the contest comes, they both drink no poison at all, and since the treasurer drank the weak poison before, he dies from it, and the pharmacist stays alive, and the king gets no poison at all.

HAHA by trying to prove, that this riddle isnt answerable in just 1 way, it hit me.
100 people drown.
for every additional boat, there are 10 more people sinking with the original 10 boats.

We can do this problem backwards. I will name it so the last person to go will be called the 1st and last is the first person to offer for simplicity.
When it's one person, he gets it all.
If there are two people, the 2nd gets it all
If there are 3 people, then the 2nd person will always disagree the split, but the 1st will take anything he can get, so the 3rd gets 99, and the first gets 1.
If there are 4 people, the 3rd will always disagree. The 4th must either offer the 2nd or the 1st more than the would have gotten in the above case, so 4th gets 99, 2nd gets 1. From here on will can assume the person that goes after the current will automatically disagree.
If there are 5 people, he must offer the 2/3 of the first 3 more than they would have gotten. So, 5th gets 98, the 3rd gets 1, the 1st gets 1.
We can continue this trend to get the answer.

Let X be the number of boats and the number of men in each boat. Then we have x^2 people to begin with. We lost 10 lifeboats, each of which has x men, so we have x^2-10x. However, we then add 10 men in each of the remaining boats, of which there are (x-10), so we have x^2 -10x +10x-100, or 100 people drown.

sry for really getting on your nerves today, but this is also wrong.
just to give you 1 example of what you havent thought of:
when 3 pirates are left, the 1. pirate will be able to get 99gold and give the last pirate 1 gold. the last pirate obv. knows that he cant get any gold from declining this offer (as you stated) so he will take the 1 gold.

ofc. this isnt the finals answer cause this way of thinking will continue to the higher first levels also.

ill will try an figure it out now

No, that's not correct. The Guru only speaks once, so he will never confirm that there are no more people with blue eyes on the island. Also there's the possibility that your eyes are not brown or blue. F.ex. your eyes could be red, you would never know since the Guru never speaks of people with red eye color.

The answer obviously lies with the Guru's statement. Since they all are perfect logicians and noone has left the island before the guru made his statement it changes everything. I don't know why though

Well, one fairly inefficient way to solve this problem is to designate counters, and use one of the switches to be a dummy switch that you switch to give no information, so the choices are basically switch or no switch on a single lightblub. The counter will just be the only person that flips the switch off, and other people would switch it on if and only if it is off and they have not been there before. It might also be possible to have multiple counters and a dynamic way of choosing counters. I don't think this is very efficient since it actually uses less possible signals then we are given.

There is one problem I can see with your answer - Won't the pirate makign the spilt jsut split it up evenly with the number of other pirates he needs to get the vote? So first step instead of 11 to all he offers 25 to four, in order to try and get the 5 votes he needs. This would mean that the 7th pirate would offer all 100 gold to the eighth pirate, and both of them would vote yes the 7th to live the 8th because he would get all the money. If i read it wrong and the person who suggests the split doesn't get to vote then the same thing would happen with the 8th offering 100g to the ninth.

The treasurer figured he had no chance to make the stronger poison, so he simply turned in something non-poisonous liquid (e.g. water), hoping the pharmacist would die from his own poison. The pharmacist realized this, so he made two poisons, and he either swallowed the weaker before the meeting, or the stronger after it. Nonetheless the king didn't necessarily get the strongest poison possible in either case.

Ah yes, I completely ignored that the treasurer wants to stay alive -_- Daemonsniper's solution is the correct one

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

-------

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

-------

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.

--------

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.

---------

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.

----------

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.

http://en.wikipedia.org/wiki/Hamiltonian_path