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OK, I have a few minutes until the pasta is ready, time for 6 and 11.
6. + Show Spoiler +OK, let's take a strict mathematical solution here:
a) If there had been one person with blue eyes, that person would only see people with brown eyes and leave on the first night.
b) Let's assume that if there were n people with blue eyes, they'd all leave on the nth night. If there were n+1 people, then each person with blue eyes would see n others with blue eyes. Since they hadn't left after n nights, therefor, they know for a fact there are n+1 people with blue eyes, and that they are one of them.
By induction, for every number of people with blue eyes, they would all leave on the night numbering the exact same amount of people since the Guru spoke.
The rest could not leave, because as far as they know, they have green or purple or whatever colored eyes. There would not be enough information for any of the rest to leave.
11. I'll probably think up an induction or proof by contradiction for it soon.
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On April 10 2011 00:56 ghrur wrote:+ Show Spoiler +9. No, it doesn't have to be prime. 31 is prime, divide by 30, remainder 1. 1 isn't prime
31 is not greater than 32.
The answer is yes because any prime number divided by a relative prime gives a remainder of another prime number
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QUESTION 5 CONCLUSIVE PROOF (BRACE YOURSELVES):
+ Show Spoiler +
I will do my best to explain this with only text, as drawings are generally necessary to convey the logical processes. It may be easier to use Proof by induction, which is also possible (I have tried both ways)
For a circles (of similar size) to actually entirely 'cover' a continuous rectangle, we note that it must intersect (overlap) with the other circles. At the extremes, it must intersect exactly at the edge of the rectangle to ensure an airtight cover. So from this we have our first postulate:
(1) All circles must intersect other circles at the edge of the rectangle (for maximum area).
Secondly, we note that rectangles can be split up into other, smaller rectangles. In fact, we can split them up such that they all become rectangles within the circles (that is, rectangles touching the edges of the circles). These squares will exist with their vertices at the intersections of the circles.
2) Rectangles can be described as a 'mosaic' of smaller rectangles.
3) Rectangles that are fully covered by circles have vertices that rest on (or inside) a circle's edges.
So from these postulates, we can bring it down to whether 4 circles of radius 0.5 can cover any rectangle that fits within a circle with a radius of 1. The result of this would be repeated as the rectangles (and overlapping circles) could then be tessellated to form some sort of rectangle. For this, we need only consider the 2 extreme cases - the square (one of the extremes) and the flat line (the other). If we draw a cartesian plane with a unit circle with center at the origin, we can see that the square is defeated by placement of the smaller circles with centers (.5,.5),(.5,-.5), (-.5,.5), (-.5,-.5). The flat line should be obvious and needs no explaining.
So from 1), 2) and 3), we deduce that if 4 small circles can cover any rectangle inside one big circle, then we can repeat the logic for any larger rectangle comprised of a multiple of these smaller rectangles.
And from above, we have shown that any rectangle in a larger circle can be covered by 4 smaller circles. So we can deduce that any rectangle made out of 25 rectangles that can each be covered by a unit circle, it can also be covered by 100 circles with radius 0.5.
PROOF BY INDUCTION (An easier explanation)
Required to prove xL<=4xS (where <= denotes that 4xS can encompass that same Area or more, than xL in relation to a rectangle as. Works the same as 'greater than or equal to' in mathematics)
Step1. It is true for x=1
L<=4S (where L is a Large circle, 4S is 4 smaller circles)
That is, what we have proven already, that a rectangle that fits in a unit circle L can be covered by 4 smaller circles S.
Step 2. Assuming it is true for x=k, prove true for x=k+1
Because 4S>=L (step 1) and 4kS>=kL (Assumption)
4S+4kS>=L+kL (Addition of equations)
therefore, 4(k+1)S>=4(k+1)L
Step 3. Therefore, because it is true for x=1, it is also true for x=1+1=2+1=3+1=4 and so on.
Therefore, it is true for x=25. 100S>=25L
Any rectangle encompassed by 25 Larger circles can be encompassed by 100 smaller circles (QED)
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On April 10 2011 01:42 Kipsate wrote:
I have 1, I found it to be a little farfetched when I knew the answer but it sticked with me due to that, I am interested to see if people can solve it:
You have 3 switches, a room and a light. 1 of these buttons turns on the light in the room, you are only allowed to go in the room once. How do you know which button turns on the light in the room?
I dislike this problem. The classical solution only works if the lightbulb is inefficient enough to generate more heat and subsequently store it in the material of the bulb such that an amount that is above the human threshold for detection remains after the convection that occurs between the time when you switch it off and walk over to it. LED's lol over this method, which might make this riddle obsolete in a few years 
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The answer is yes because any prime number divided by a relative prime gives a remainder of another prime number
The remainder of 109 divided by 60 is 49, which isn't prime.
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On April 10 2011 03:18 gyth wrote:Show nested quote +The answer is yes because any prime number divided by a relative prime gives a remainder of another prime number The remainder of 109 divided by 60 is 49, which isn't prime.
It's because no odd composite number between 2 and 30 fails to have a common denominator with 30. It is impossible to form a composite number with a remainder of 9, 15, 21, 25, or 27 after dividing it by 30.
It's also difficult to understand your latest problem:
15. Let A be a collection of 100 distinct integers. Can you select 15 integers from A so that the difference of any two numbers from this selected subset is divisible by 7?
Which 100 distinct integers? It's theoretically possible to have a collection of 100 distinct integers in which this kind of selection is impossible (100 multiples of 13, for instance. Only multiples of 91 are then selectable.)
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7. (accessible to everyone) Suppose we have 9 coins that look the same and feel the same. But exactly one of them is counterfeit and it weighs less than a real coin. Can we identify the counterfeit coin among the 9 coins with just two weighings on an accurate balance scale?
Answer:
+ Show Spoiler +Assume coins A, B, C, D, E, F, G, H, I
Place A B C in one side of the scale and D E F in the other.
1) A B C < D E F
Means, that either A, B or C is counterfeit
Compare A and B in the scale
1.A) A < B
A is light.
1.B) A > B
B is light.
1.C) A = B
C is light.
2) A B C > D E F
Means, that either D, E or F is counterfeit
Compare D and E
2.D) D < E
D is light.
2.E) D > E
E is light.
2.F) D = E
F is light.
3) A B C = D E F
Means, that either G, H or I is counterfeit
Compare G and H
3.G) G < H
G is light.
3.H) G > H
H is light.
3.I) G = H
I is light.
GG
8) If you have 2 pieces of string that when you light in fire take an hour to burn how do you measure 45 minutes?
Answer:
+ Show Spoiler +
1) Light stick 1 in one end.
2) Light stick 2 in both ends.
3) When stick 2 has finished burning, it has passed 30 minutes. So stick 1 will burn for 30 minutes more.
4) Light stick 1 in the other end = 15 minutes
Total time: 45 minutes.
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New one:
People are seated around the round table in regular intervals (spaced equally to fill entire table). If 9th person is sitting directly across 22nd, how many people are seated at the table?
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Which 100 distinct integers? It's theoretically possible to have a collection of 100 distinct integers in which this kind of selection is impossible (100 multiples of 13, for instance. Only multiples of 91 are then selectable.)
If you had the first 99 multiples of 13, then 15 of them would have differences of 91.
(14*7+1=99)
Not sure what language to use to go about proving that though.
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On April 10 2011 04:04 Manit0u wrote:
People are seated around the round table in regular intervals (spaced equally to fill entire table). If 9th person is sitting directly across 22nd, how many people are seated at the table?
+ Show Spoiler +
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It's a convoluted variation of the monty hall problem. I like this one.
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For #1, didnt have time to look at others
+ Show Spoiler +
#1, add Fill 3L Jug, pour all in 5L jug,
#2, Repeat Step 1
#3, 1L excess falls out
#4 Repeat 1,2,3 Till you have 4L
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I think I've got solutions for #1, #2 and #2b.
+ Show Spoiler +
#1:
Fill up the 5 liter jar completely. Pour from 5l to 3l until 3l is full. You now have 2l left in the 5l jug.
Empty 3l jug. Put remaining 2l in the 3l jug. Fill up 5l jug. You now have 2 liters in the 3l jug, and 5 liters in the 5l jug. Now just fill water from the 5l into the 3l jug until the 3l jug is full. You now have 4l in the 5l jug.
#2 Yes. If it is a perfect scale, it is precise enough to determine the difference in the gravitational force that acts between the earth and the bags. I'm not going to give you the exact equations now, but I can describe it: You stack the bags ontop of each other, and weigh them. Each configuration is going to correspond to one number showing up on the scale (the highest being the one when the 0.9x is on the top of the stack, the lowest one being when the 0.9x bag is at the bottom of the stack), which you can use to determine where the bag is in the stack.
#2b Yes. You again stack them up and weigh them all together. Using the same principle as above, the several possible combinations will correspond to a certain set of numbers which show up on the scale. gg
Will look at 3 later.
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On April 10 2011 01:35 Slunk wrote:Show nested quote +On April 10 2011 01:21 Starfox wrote:On April 10 2011 01:08 Slunk wrote:
Question to #6
Is it correct to assume that the guru is completely useless and only built in as a distraction? Nope, the guru plays a roll in the solution, it's a rather famous riddle, the numbers could be anything though, the logical conclusion goes for every number of people ]But the guru only speaks once and tells something that everybody knows allready.
6. + Show Spoiler +All of them leave on the first night! The solution is pretty simple and the guru actually is pretty useless, ah all he does is to confirm, that the solution is possible. I really don't want to spoil it, so click it, only if you are hundred percent sure you want to know it. + Show Spoiler [solution] +Two random people will stand next to each other. All other guys on the islands (save the guru - he already know his color (he DOES know he is the guru, since he executes the role)) will stand in a column in front of them. 1. First guy in the column will go to the first guys. 1a. If they have the the same eye color, he will go to one of their sides. Doesn't matter left or right. 1b. If they have different eyes colors he will go to the middle between them, hence there will be clear "middle" point between the half-row with blue eyes and half-row with brown eyes. 2. Second guy and so forth, save the last one will repeat step 1. This way the one that goes in the "middle" point always sees where people in front of him divide by eye color and move between them, creating new such point. 3. The last guy will have to count the number of people with blue eyes and people with brown eyes in front of him (We shall assume he has not done so yet). The statement of the tasks clearly says:
On this island there are 100 blue-eyed people, 100 brown-eyed people, and the Guru.
By counting which group in front of him has 99 people, he will know to which group he belong and move to the side of corresponding half-row. By doing so, all the rest 99 people will know the color of their eyes, as they will see the color of last person eyes when he moves. The other half-row will also know, they have the opposite color of the last man. Well, now everybody knows the color of his (hers with the guru) and all 201 people will leave the island on the first night. You might need to visualize this to understand how it works, but it's pretty simple and straightforward. It's actually harder to explain all the steps.
Do i get a virtual present now?
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On April 10 2011 01:42 Kipsate wrote:
I have 1, I found it to be a little farfetched when I knew the answer but it sticked with me due to that, I am interested to see if people can solve it:
You have 3 switches, a room and a light. 1 of these buttons turns on the light in the room, you are only allowed to go in the room once. How do you know which button turns on the light in the room?
Answer:
+ Show Spoiler +
Turn on switch #1 on and leave it so for some time - say 10 minutes.
Turn it off and turn on switch #2.
Go in the other room.
If the bulb is on, is connected to switch #2. If it is off, but it's cold, then it's connected to switch #3.
If it is off, but it's hot (when touched), then that's it's indicator, the bulb has been on a bit ago, hence it is connected to switch #1.
Standard "think outside of the box".  I want a teddy bear now...at least virtual one.
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Yay! TSL is over, and there is another question!
15. + Show Spoiler +If you have 7 * (1 - 1) + 2 distinct numbers, you can select 1 so that any choice of 2 selected numbers will have a difference divisible by 7.
Let's assume that if you have 7 * (n - 1) +2 distinct integers which if you select n of them, the difference of any 2 of those n can be divided by 7.
If you have 7 * n + 2 such integers, you can choose 7 * (n-1) + 2 of them so that out of the remaining 7, you cannot select 2 in which their difference will be divisible by 7 (this is possible, because if it wasn't, you could choose 2 or more out of them, and then would be able to choose less from the rest with ease). Therefor, you can select n from the 7*(n-1) + 2 of them, and you need to select one of the remaining 7. As they are of the forms 7a, 7b+1, 7c+2, 7d+3, 7e+4, 7f+5, 7g+6, at least one of them will be selectable so that the difference between it and the rest will be divisible by 7.
By induction, for any natural number n, if you have a group of 7*(n-1)+2 distinct integers, you can select n of them so that the difference between any two will be divisible by 7.
And specifically, for a set of 7 * (15 - 1) + 2 (which equals 100) distinct integers, you can select 15 of them so that the difference between any two will be divisible by 7.
edit: ooh, there's 2b!
2b. + Show Spoiler +You take one from the first bag, 2 from the second, 4 from the third and 8 from the fourth. Therefor, the weight will be 15 - 0.1n grams, n being the number of fake coins. switch n to a binary representation. It will be a 4 digit binary number, and from left to right, if there is a one in that place, the 4th, 3rd, 2nd, or 1st bag will be of fake coins.
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15.+ Show Spoiler +If we group the 100 numbers by their mod 7. The difference between members in a group would be a multiple of 7. By pigeon hole, the smallest the largest group could be is 15 members. Why +2? Wouldn't you only need a 99 member starting group?
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Answer to 8
+ Show Spoiler +Light both ends of one rope and one end of the other.
The rope lit at both ends will take half an hour to burn down, when this happens light the other end of the remaining rope which will now take fifteen minutes to burn out.
Hey Presto, 45 minutes =D
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EPT
