Naked Science Forum
General Science => General Science => Topic started by: Alan McDougall on 30/06/2008 19:39:16
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Hello,
A little logic problem I hope it is posted in the correct place, If not please move it
x: I forgot how old your three kids are.
y: The product of their ages is 36.
x: I still don't know their ages.
y: The sum of their ages is the same as your house number.
x: I still don't know their ages.
y: The oldest one has red hair.
x: Now I know their ages!
How old are they
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There are two twins aged 2 and the eldest is 9.
Bit of a chestnut, eh?
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Blake,
From the statement that the product of their ages is 36 the possibilities of the three individual ages are:
1,1,36
1,2,18
1,3,12
1,4,9
1,6,6
2,2,9
2,3,6
3,3,4
From the statement that the sum equals the house number it is possible to eliminate all but two possibilities. The sums of the rest are unique and would allow for an immediate answer. For example if the house number were 16 the ages must be 1, 3, and 12. The two remaining possibilities are 2, 2, and 9; or 1, 6, and 6.
After the clue that the oldest has red hair you can eliminate 1, 6, and 6 because the oldest two have the same age thus there is no oldest son. The only remaining posibility is 2, 2, and 9.
Answer
Their ages are 2, 2, and 9
You were correct maybe next time something much more difficult.
Do you know the 12 ball oddball problem?
Alan
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Alan what do you do if the family concerned might live in their home, that only has a name on it's address not a number in it's roadway??
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Rosalin,
Alan what do you do if the family concerned might live in their home, that only has a name on it's address not a number in it's roadway??
You will not be able to solve the problem in the way I suggested, let me think about this and come back to you
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Do you know the 12 ball oddball problem?
That is?
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OK
You notice I have changed thread title
Here are three more easy problems before we try something really difficult
try these easy tests of logical thinking.
Their are twelve identical featureless balls of equal volume and size. One differs minutely in mass or weight if you like.
The task is to establish in three weighing steps using a very sensitive balance scale, like the scale of Libra. Which ball is different and if it is heaver or lighter. You can use any combination , three and three, four against four, anything you like but you must solve the problem in three weighs. You cant use a bathroom scale this would be useless.
O O O O O O O O O O O O
Problem two
A person is rowing his bout upstream in a river flowing at three miles an hour at seven miles per hour relative, to the bank of the river. His hat falls off and only after 45 minutes does he notice this. He immediately turns around and rows at the same speed to get his beloved hat. (disregard the time taken for turning around for the purpose of this test).
How long does it take for him to catch up to his hat and retrieve it??
Last problem
A man sentenced to death is given a choice. He is put in a room with two PC computers, one is programed to only lie, and the other programed to only tell the truth. There are two exit doors , one leading to the death chamber and the other to freedom. He is only allowed to key in "one question" to only "one of the computers" and by this one question, he must establish the door to freedom or face execution.
Give it a go
alan
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OK
You notice I have changed thread title
Here are three more easy problems before we try something really difficult
try these easy tests of logical thinking.
Their are twelve identical featureless balls of equal volume and size. One differs minutely in mass or weight if you like.
The task is to establish in three weighing steps using a very sensitive balance scale, like the scale of Libra. Which ball is
different and if it is heaver or lighter. You can use any combination , three and three, four against four, anything you like
but you must solve the problem in three weighs. You cant use a bathroom scale this would be useless.
O O O O O O O O O O O O
3 groups of 4 balls; one group in a plate of the scale, another in the second plate; this way we identify the group with the lighter ball (if the 2 groups in the scale have the same weight, then the third group is the one we are looking for). We divide this group in two sub-groups of 2 balls and we put them on the scale, so we identify the subgroup with the lighter ball. With the last measure (one ball in each plate) we identify the single ball.
Problem two
A person is rowing his bout upstream in a river flowing at three miles an hour at seven miles per hour relative, to the bank
of the river. His hat falls off and only after 45 minutes does he notice this. He immediately turns around and rows at the
same speed to get his beloved hat. (disregard the time taken for turning around for the purpose of this test).
How long does it take for him to catch up to his hat and retrieve it??
If I understood the english correctly, he goes at 3 miles/h with respect to the river and the river flows at 4 miles/h. In a ref. frame still with respect to the river, he looses the hat, goes ahead for 45 minutes and then turns back. Of course he'll take the same 45 minutes to return to the point he lost the hat regardless of his speed and on the river's speed.
Last problem
A man sentenced to death is given a choice. He is put in a room with two PC computers, one is programed to only lie, and the
other programed to only tell the truth. There are two exit doors , one leading to the death chamber and the other to freedom.
He is only allowed to key in "one question" to only "one of the computers" and by this one question, he must establish the
door to freedom or face execution.
"If I asked the other PC which door leads to the death chamber, what did it answer?" The answer is the door to freedom.
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3 groups of 4 balls; one group in a plate of the scale, another in the second plate; this way we identify the group with the lighter ball (if the 2 groups in the scale have the same weight, then the third group is the one we are looking for). We divide this group in two sub-groups of 2 balls and we put them on the scale, so we identify the subgroup with the lighter ball. With the last measure (one ball in each plate) we identify the single ball.
How does this work if we don't know whether the odd ball is lighter or heavier? Surely, we would need to add an extra weighing stage to work this out - ie group A is lighter than group B - this is either because there's a light ball in A or a heavy ball in B. We then need to test one against group C - If it's equal to B, the 'odd' ball is light, if it's equal to A the 'odd' ball is heavy. But that's used up one of our tests, and now we only have one left to determine which one of the 4 it could be.
If we get lucky and A & B are identical - we know that the odd ball is in group C. So we split C into C1 and C2, and C1 comes out lighter - we split that and find they're equal - we then know that the odd ball is one of C2, and that it's heavy, but we've run out of tests.
This can't just rely on getting lucky, can it?
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3 groups of 4 balls; one group in a plate of the scale, another in the second plate; this way we identify the group with the lighter ball (if the 2 groups in the scale have the same weight, then the third group is the one we are looking for). We divide this group in two sub-groups of 2 balls and we put them on the scale, so we identify the subgroup with the lighter ball. With the last measure (one ball in each plate) we identify the single ball.
How does this work if we don't know whether the odd ball is lighter or heavier?
Right, I lost the part where he says we don't know if it's lighter or heavier... [:I] Surely, we would need to add an extra weighing stage to work this out - ie group A is lighter than group B - this is either because there's a light ball in A or a heavy ball in B. We then need to test one against group C - If it's equal to B, the 'odd' ball is light, if it's equal to A the 'odd' ball is heavy. But that's used up one of our tests, and now we only have one left to determine which one of the 4 it could be.
Maybe we could do this way, but don't know if it's considered one more measure (for a total of 3) or two more measures: when we have identified the odd group of 4, let's say we have found it's lighter, we put 1 ball in each plate; if we are lucky and one is lighter, it's finished; if they weight the same, we add another ball to each plate so we discover which of the last two is lighter. (I know, it seems cheating... [:)])
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I have seen this before without the heavier/lighter thing, so I think we can let you off for missing that bit!
It does seem like cheating - and this sort of thing shouldn't rely on luck.
Come on then Alan - how does this one work?
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Lightarrow,
Respectfully you are incorrect you assumed the odd ball was lighter, it could be lighter or heavier, the problem is more complex than your solution.
Ben
Luck must play no part in the solution it must be fool proof (no pun intended)
Regards
Alan
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Lightarrow,
Respectfully you are incorrect you assumed the odd ball was lighter, it could be lighter or heavier, the problem is more complex than your solution.
Ben
Luck must play no part in the solution it must be fool proof (no pun intended)
Regards
Alan
Thank you for the "Respectfully". I wasn't able to find the solution. Can you give us the answer?
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Me too - I can solve it in two weighings if I cheat and use either gravitational attraction or momentum, but not purely via weighing.
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Good morning people,
Here is the solution(s) to the 12 odd ball problem!
You could use chalk to identify ball potially lighter or heaver or both
Before I explain some conventions in setting out the solution to the Twelve Balls problem, here again is a statement of the task.
There are twelve balls, all the same size, shape and color. All weigh the same, except that one ball is minutely different in weight, but not noticeably so in the hand. Moreover, the odd ball might be lighter or heavier than the others.
Your challenge was to discover the odd ball and whether it is lighter or heavier.
You must use a beam balance only, and you are restricted to three weighing operations.
Note: by starting with 6 /6 or, 5/ 5 or 3/ 3. or 2/2 or 1/1 it is impossible to solve the problem
“The only way to solve this rather complex problems is by starting by weighing 4 balls against 4 as you will see by the solution below”
Conventions:
At every weighing one of three things theoretically can happen: the pans can balance, the left pan can go down or the left pan can go up.
It will be necessary to refer to a given ball as definitely normal (N), potentially “heavier” (H) or potentially “lighter” (L). Often our identification of a ball in this way will be as part of a group (= “This group contains a heavier/lighter ball”), and will depend on what we learn from a previous weighing. At the start, all balls have a status of unknown (U).
To show at each weighing what is being placed in each pan, represent the situation as per the following examples:
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First Weighing UUUU ——— UUUU
Pans balance All these U’s are now known to be N’s; the odd ball is one of the remaining unweighed four (call them UUUU from now on).
Proceed to Second Weighing: Case 1
Left pan down
One of the four balls in the left pan might be heavier (call them HHHH from now on) or one of the four balls in the right pan might be lighter (call them LLLL from now on).
Proceed to Second Weighing — Case 2
Left pan up
One of the four balls in the left pan might be lighter (call them LLLL from now on) or one of the four balls in the right pan is heavier (call them HHHH from now on).
Proceed to Second Weighing — Case 2
Second Weighing
Case 1 UUU ——— NNN
Pans balance All these U’s are now known to be N’s; the odd ball is the remaining unweighed U, but we don’t yet know if it’s heavier or lighter than normal.
Proceed to Third Weighing — Case 1
Left pan down
One of these U’s is heavier than normal, but we don’t yet know which one (call them HHH from now on).
Proceed to Third Weighing — Case 2
Left pan up
One of these U’s is lighter than normal, but we don’t yet know which one (call them LLL from now on).
Proceed to Third Weighing — Case 3
Case 2 HHL ——— HLN
Pans balance
All these H’s and L’s are now known to be N’s; the odd ball is one of the remaining unweighed H or two L’s.
Proceed to Third Weighing — Case 4
Left pan down
The odd ball is one of the left two H’s or the right L.
Proceed to Third Weighing — Case 5
Left pan up
The odd ball is either the right H or the left L.
Proceed to Third Weighing — Case 6
Third Weighing
Case 1 U ——— N
Pans balance Not possible
Left pan down The odd ball is this U, and it’s heavier
Left pan up The odd ball is this U, and it’s lighter
Case 2 H ——— H
Pans balance The odd ball is the remaining unweighed H (heavier)
Left pan down The odd ball is the left H (heavier)
Left pan up The odd ball is the right H (heavier)
Case 3 L ——— L
Pans balance The odd ball is the remaining unweighed L (lighter)
Left pan down The odd ball is the right L (lighter)
Left pan up The odd ball is the left L (lighter)
Case 4 L ——— L
Pans balance The odd ball is the remaining unweighed H (heavier)
Left pan down The odd ball is the right L (lighter)
Left pan up The odd ball is the left L (lighter)
Case 5 H ——— H
Pans balance The odd ball is the remaining unweighed L (lighter)
Left pan down The odd ball is the left H (heavier)
Left pan up The odd ball is the right H (heavier)
Case 6 H ——— N
Pans balance The odd ball is the remaining unweighed L (lighter)
Left pan down The odd ball is this H (heavier)
Left pan up Not possible
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First Weighing UUUU ——— UUUU
...
Left pan down
One of the four balls in the left pan might be heavier (call them HHHH from now on) or one of the four balls in the right pan might be lighter (call them LLLL from now on).
Proceed to Second Weighing — Case 2
...
Second Weighing
...
Case 2 HHL ——— HLN
Where did you take that "N" ball?
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Where did you take that "N" ball?
I assume from where did Get the N ball, I got it from the unweighed group of 4 balls that I knew were normal n balls. After using it I put it aside.
Cut out 12 peaces of paper and imagine you are doing the problem on an imaginary scale. You will see the solution I gave is the correct one.
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Where did you take that "N" ball?
I assume from where did Get the N ball, I got it from the unweighed group of 4 balls that I knew were normal n balls. After using it I put it aside.
Right, thank you.
Very interesting problem! Thank you for posting it. I'm waiting for the next ones!
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Oh by the way people I said the problems were easy and this is not true about the odd ball problem.
I was given that problem over 30 years ago my a work colleague (and solved it myself).
I have posed the problem to numerous People since then and only one person in all that time was able to give the correct answer.
If you did not solve it believe me you are in good company.
Regards
Alan
I will think of another and come back
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Just to confuse people further. Did you know it's possible to answer the question about the balls without any weighings i.e. without the scales or even in zero gravity?
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Just to confuse people further. Did you know it's possible to answer the question about the balls without any weighings i.e. without the scales or even in zero gravity?
Making a ball at rest collide with another ball and see if this last one stops or goes ahead or goes back after collision?
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Yes, basically play snooker with them and the odd one should show up.
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Yes, basically play snooker with them and the odd one should show up.
Ok, then, for the others readers I explain:
If you collide a ball A at rest with another ball B of the same dimension and mass (the two balls' centers of mass aligned with the trajectory) then after collision B ball stops completely; if instead the B ball were lighter, after collision it bounce back (slower); if it were heavier, it keeps going in the same sense (slower).
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LA,
Correct but it might be difficult with balls that differ minutely.
And it would not solve the 12 ball problem .
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"Correct but it might be difficult with balls that differ minutely."
Same as a set of scales then.
"And it would not solve the 12 ball problem ."
Yes it does.
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"No it does not" If luck is not on your side your solution might require as much as 36 bouncing combinations before finding the odd ball.
I am aware the less massive ball will bounce back. In my youth I also played snooker.
Alan
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"No it does not" If luck is not on your side your solution might require as much as 36 bouncing combinations before finding the odd ball.
Actually, with only one shot, if you put all the balls in a row, quite distant one another, and you shoot the first (or the last). [:)]
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Sorry it makes no sense!
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Sorry it makes no sense!
Why?
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It still makes no sense you are only allowed three weights/bounced/shots
Shots
If I understand you correctly, you want to bounce the balls against each other,hopefully by careful observation you would see the ball with less mass bounce futher back than the one with the greater mass.
To do this you can choose and ball and bounce it against any other ball. If luck is not on your side it would need 11 bounces to slove the problem. Not three weighs on a pivot scale as in my 12 odd ball solution
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The whole point is that this way I need precisely zero weighings, I might need a dozen shots or so (I'm alousy player) but nobody said anything about them in the original question. In principle I could, as Lightarrow says, do it with one shot.I could do this in zero gravity with no weighings so that's less than three or do you somehow think 3 is not more than zero?
Let's make this as clear as I can.
I can solve the problem of finding the odd ball and whether it's light or heavy with zero weighings.
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OH!!Man come on!! be real, your experiment solution would require a trip the the space atation. It is not practicle and completely irrelevant to my 12 all problem.
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Alan, can you post the other problems you said to have? I find them interesting.
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lightarrow,
Ok here is three more, I hope they are of interest!
Try these three problems and if they are found of interest I will pose more
1) Two brothers share a flock of x sheep. They take the sheep to the market and sell each sheep for $x. At the end of the day they put the money from the sales on the table to divide it equally. All money is in $10 bills, except for less than ten excess $1 bills. One at a time they take out $10 bills. The brother who draws first also draws last. The second brother complains about getting one less $10 bill so the first brother offers him all the $1 bills. The second brother still received a total less than the first brother so he asks the first brother to write him a check to balance the things out. How much was the check
2) Four dogs occupy the four corners of a square with side of length a. At the same time each dog starts walking at the same speed directly toward the dog on his left. Eventually all four dogs will converge at the center of the square. What path does each dog follow and what is the distance each dog walks until he reaches the center?
3) Five pumpkins are weighed two at a time in all ten sets of two. The weights are recorded as 16, 18,19, 20, 21, 22, 23, 24, 26, and 27 pounds. All individual weights are also integers. How much does each pumpkin weigh?
Some of the problems are from internet, most are mine collected over the years. I love problems of deductive logical thinking.
We can go the route of the really difficult but that would be no fun I think
Regards
Alan
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"OH!!Man come on!! be real, your experiment solution would require a trip the the space atation. It is not practicle and completely irrelevant to my 12 all problem."
Nonsense.
At most, it would require a trip to the local pool hall. In fact anywhere with a reasonably flat table or floor would do.
The fact is that, if you happened to be on a space station, my method would still work, but your's wouldn't.
It's perfectly practical; what's impractical about playing snooker or pool?
Your original question was just a puzzle- any solution that meets the stated objective is valid. In terms of minimising the number of weighings, my solution beats yours.
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Yes I wanted to say it could be done "your way" on earth. But your suggestion has nothing to do with the puzzle. To say anything that meets the objective is silly and outside the confines of the odd ball problem.
Regards
Alan
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2) Four dogs occupy the four corners of a square with side of length a. At the same time each dog starts walking at the same speed directly toward the dog on his left. Eventually all four dogs will converge at the center of the square. What path does each dog follow and what is the distance each dog walks until he reaches the center?
The distance walked is a. The path is an exponential spiral with respect to the square's center (a/2;a/2):
r(t) = -a/2 exp[-θ(t)]*{cos[θ(t)] + sin[θ(t)];cos[θ(t)] - sin[θ(t)]} + (a/2;a/2)
|r(t) - (a/2;a/2)| = a/sqrt(2) exp[-θ(t)]
About the distance walked my reasoning is probably not simple:
after a time dt the dog (I considered the one at the left down corner in the origin of coordinates) walks for vdt up and vdt at right so he is now at a distance sqrt[(a-vdt)^2 + (vdt)^2] ~ a - vdt so it's now in a corner of a new square whith side of lenght: a - vdt; so the differential variation of the square's side lenght is da = (a - vdt) - a = -vdt and this equation is valid in every moment. Integrating from 0 to t:
∫da = ∫-vdt --> a(t) - a = -vt --> a(t) = a - vt
When they meet, at the instant T, they are in a square of side lenght 0, so:
0 = a - vT --> T = a/v
Since the speed v is constant, the total lenght L is simply v*T:
L = v*a/v = a.
About the path followed, my reasoning was even less simple, because I solved it analitically (complicated computations), so I assume there must be a simpler way.
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"To say anything that meets the objective is silly and outside the confines of the odd ball problem."
Yet it still solves the problem of spotting the oddball, and it does so with fewer weighings.
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Lightarrow,
Lightarrow,
I like you approach to this not so simple little problem
Hard Solution
It is intuitive that at all times the dogs will form the corners of a square, ever decreasing in size and rotating about the center.
Let C be the center of the square at point (0,0).
Let the dogs be at initial points (.5,.5), (.5,-.5), (-.5,-.5), and (-.5,.5).
Let dog 1 bet at (.5,-.5) who is chasing dog 2 at (.5,.5). Consider the line from C to dog 1.
Next consider the line from dog 1 to dog 2. The angle between these lines is 135 degrees.
Theorem: Let C be a fixed point, D a point on a curve, r the distance from C to D, x the polar angle formed by r, and y the angle formed by the angle between CD and the tangent line to the curve at point D. Then tan(y)=r/(dr/dx).
So the line of sight from the center to the initial point of dog 1 would be be at a 270 degree angle. The line of sight from dog 1 to dog 2 would be a vertical line of 90 degrees. The angle between these lines is 270 degrees. The tangent of 270 is -1. So we have r/(dr/dx) = -1. Thus r(x)=c*e-x.
At the moment the dogs start walking x equals 0. At this moment the distance from the center to any of the dogs is 2-1/2. So r(x)=2-1/2*e-x.
The arc length is given by the formula:
Integral from 0 to infinity of ((f(x))2+(f'(x))2)1/2 =
Integral from 0 to infinity of ((2-1/2*e-x)2 + (-2-1/2*e-x)2)1/2 =
Integral from 0 to infinity of (2-1*e-2x + 2-1*e-2x)1/2 =
Integral from 0 to infinity of e-x =
-e-x from 0 to infinity =
=0 - (-e0) =
0 -(-1) = 1 The path the dogs take is called a logarithmic spiral.
It is an interesting paradox that the dogs will make an infinite number of circles, yet the total distance is constant. Every revolution the size of the square will to e-2*pi/21/2 = 0.03055663 times it's size before the revolution.
Easy Solution
David Wilson suggested the more simple solution which follows...
Suppose dog A is pursuing dog B who is pursuing dog C. During the entire pursuit, the dogs remain at the corners of a square, and angle ABC is a constant 90 degrees. That is, B's path towards C is always perpendicular to A's path towards B, and B has zero velocity in the direction along A's path. Since B neither approaches nor recedes from A during the walk, A simply covers the intial separation of 1.
Antonio Molins went further to state that as dog A is heading direcctly towards dog B, dog B is moving towards from dog A at rate equal to cosine of any of the interior angles of the shape made by the dogs.
He also provided the following graphics.
For example consider a pentagon. Remember that for a n-gon of n sides each angle will be 180*(1-(2/n)).
So a pentagon will have angles of 108°. Let's assume each dog walks at a rate of 1 unit per minute, where the initial shape has sides of 1 unit. As pointed out by David Wilson we can forget about the spiral paths but instead think of the dogs walking in straight lines.
In the case of the penagon for every one unit dog A draw towards dog B's initial position, dog B will draw cos(180°)=-0.309 units towards dog A.
The question is, how long will it take for them to meet? Let the answer to that question be t. In t seconds dog A will cover t units, and dog B will cover .309*t units. The total distance traveled by A equals the distance traveled by B plus 1 (the initial separation). So solving for t: t=0.309+1 --> t=1/(1-.309) --> t=1.4472.
What do you think?
Alan
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OK,
How about this little puzzle.
A horse is tied to a stake with a rope 5 metres in length, 10 metres in front of him is a bunch of nice hay. The hay is firmly tied down and can not be moved.
The horse somehow gets to the hay and begins to eat. How does he do it?
Regards
Alan
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Now try this
Please don't muse over the request act immediately you read what you must do
Before anything get a pencil and paper ready. please don't scroll down yet
OK scroll down now:.................? far down please!!
Try to think of two objects or shapes "one inside the other" under the circumstances I suggested and "IMMEDIATELY DRAW THEM"
as soon as the image pops into your mind.
Note don't reveal what you drew I will attempt to predict that later, after a few replies, such as OK Alan I have my drawing!!
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OK,
How about this little puzzle.
A horse is tied to a stake with a rope 5 metres in length, 10 metres in front of him is a bunch of nice hay. The hay is firmly tied down and can not be moved.
The horse somehow gets to the hay and begins to eat. How does he do it?
Regards
Alan
Forgive me if I'm being too logical about this one, but horses are pretty good at chewing through rope. If was a chain, I'd have to think about it again...
Oh, and I've thought of a shape within a shape.
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Perhaps someone takes pity on him?
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No no guys with repect there is a more logical answer. The horse must get to the hay all on his own, no outside help permitted
Regards
Alan
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So "he chews through the rope" doesn't score any points?
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I don't know how to hide an answer, but I doubt this is the answer being looked for.
The puzzle states that the hay is tied down. It does not state that the stake is driven into the ground so maybe the horse just walks over to the hay dragging the stake.(Though my first thought was that he chews through, or breaks the rope.)
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OK,
How about this little puzzle.
A horse is tied to a stake with a rope 5 metres in length, 10 metres in front of him is a bunch of nice hay. The hay is firmly tied down and can not be moved.
The horse somehow gets to the hay and begins to eat. How does he do it?
Regards
Alan
The horse is standing at the extreme length of the rope, five meters from the stake. The horse is facing the stake, and five meters further away in the same direction is the hay. The horse walks the ten meters across the 'circle' and eats the hay.
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BoardChemist,
I don't know how to hide an answer, but I doubt this is the answer being looked for
.
"" to the mind guess"" of prediction test in "blue letters", This is post separate from the horse hay one.
The puzzle states that the hay is tied down. It does not state that the stake is driven into the ground so maybe the horse just walks over to the hay dragging the stake.(Though my first thought
was that he chews through, or breaks the rope.)
Yessssssss!! The above is the correct answer a little silly I know but many people most often just presume the skate is firmly hamered into the ground
Regards
Alan
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I'm fairly sure I know the answer to the pumpkins one, just haven't yet given myself the time to write it out! Back with you soon on that... and knowing horses, a horse would have chewed through the rope anyway, even if the stake wasn't in the ground. They just like to be a pain.
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Lost track of all that..
1)Weigh 4 against 4.. Either balance or not.
2)Swap 1 from one pan to the other... Replace 3 balls with 3 un-weighed balls.
Lock at change of state between first and second weighing. In all cases you can find the odd one.. Might have to indentify one as a control.
It come down to one of two balls, lighter or heavier... Weigh one against a control.
OR one of the three where you now know you are looking for a lighter or heavier. Weigh any two together.
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I suppose it could have been a sawing horse - so no problem, again.
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The horse was standing next to some hay in the first place, there were two bundles, but where were the oats? Did the horse have his oats?
Ok, I have thought of a picture.
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Two glasses, one nearly filled with liquid A, one with liquid B. Take a teaspoon of A and put it in B, stir and put a teaspoon of the mixture back in A. Which now is the purest??
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Okay, lets get back to the pumpkins...
Five pumpkins are weighed two at a time in all ten sets of two. The weights are recorded as 16, 18,19, 20, 21, 22, 23, 24, 26, and 27 pounds. All individual weights are also integers. How much does each pumpkin weigh?
So, there are 5 pumpkins - lets call them a, b, c, d and e, and assume they are in ascending weight order.
And they're weighed in pairs:
a+b, a+c, a+d, a+e, b+c, b+d, b+e, c+d, c+e and d+e
This means that each pumpkin was weighed 4 times, and so the total of all the weighings/4 is the total weight of all 5 pumpkins
16+18+19+20+21+22+23+24+26+27 = 216 ... 216/4 = 54lbs
Now, if a and b are the smallest, then a+b = 16lbs
If d and e are the largest, then d+e = 27 lbs
So 54lb - (a+b)+(d+e) = c
54 - (16+27) = 11
So pumpkin c weighs 11lb.
And from there, it should be relatively easy to work out all the other weights:
a+b = 16
a+c = 18 (and therefore a= 7lb, and b=9lb)
a+d = 19 (and therefore d= 12lb)
...
d+e = 27 (and therefore e=15lb)
So:
a= 7lb
b= 9lb
c= 11lb
d= 12lb
e= 15lb
How's that?
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Two glasses, one nearly filled with liquid A, one with liquid B. Take a teaspoon of A and put it in B, stir and put a teaspoon of the mixture back in A. Which now is the purest??
I'm going straight for the obvious answer(s), which may land me in trouble, but:
i - After all the stirring, you would have glass A with just under a teaspoon of B in, and glass B with just under a teaspoon of A in, so they're both pretty much equal
ii - neither of them are pure anymore...
Ah Ha! I've spotted the trap! You're adding 5ml of A to a full glass of B, then 5ml of B(+A) to a glass of A minus 5 ml. So is B more pure?
I've lost my train of thought now...
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It is not B.
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It's not A either. (I'm assuming that the volumes of the 2 liquids were identical originally.)
Unless, of course, something odd happens.
There's the complication that mixing liquids can give rise to a change in volume. Mixing a litre of alcohol with a litre of water doesn't give 2 litres of the mixture.
Otherwise, given that both glasses start and end with the same volume and that whatever is gained by one must have been lost from the other the concentration aof A in B must be the same as the concentration of B in A.
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Benv,
So, there are 5 pumpkins - lets call them a, b, c, d and e, and assume they are in ascending weight order.
And they're weighed in pairs:
a+b, a+c, a+d, a+e, b+c, b+d, b+e, c+d, c+e and d+e
This means that each pumpkin was weighed 4 times, and so the total of all the weighings/4 is the total weight of all 5 pumpkins
16+18+19+20+21+22+23+24+26+27 = 216 ... 216/4 = 54lbs
Now, if a and b are the smallest, then a+b = 16lbs
If d and e are the largest, then d+e = 27 lbs
So 54lb - (a+b)+(d+e) = c
54 - (16+27) = 11
So pumpkin c weighs 11lb.
And from there, it should be relatively easy to work out all the other weights:
a+b = 16
a+c = 18 (and therefore a= 7lb, and b=9lb)
a+d = 19 (and therefore d= 12lb)
...
d+e = 27 (and therefore e=15lb)
So:
a= 7lb
b= 9lb
c= 11lb
d= 12lb
e= 15lb
"Correct and well thought out Benv"
Answer
Five pumpkins are weighed two at a time in all ten sets of two. The weights are recorded as 16, 18,19, 20, 21, 22, 23, 24, 26, and 27 pounds. All individual weights are also integers. How much does each pumpkin weigh
Problem 162 Answer
The answer is 7, 9, 11, 12, and 15 pounds.
Thanks to Elizabeth Gomez for this problem.
And thank BenV
Regards
Alan
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Some more problems that I think are more fun to solve!!
1) The Three Suspects
Kyle, Neal, and Grant were rounded up by their mother yesterday, because one of them was suspected of having grabbed a few too many cookies from the cookie jar. The three brothers made the following statements under very intensive questioning:
• Kyle: I'm innocent.
• Neal: I'm innocent.
• Grant: Neal is the guilty one.
If only one of these statements was true, who took the cookies?
2) The Three Switches
In your basement are three light switches, all of them currently in the OFF position. Each switch controls one of three different lamps on the floor above.
You would like to find out which light switch corresponds to which lamp.
You may move turn on any of the switches any number of times, but you may only go upstairs to inspect the lamps just once.
How can you determine the switch for each lamp with just one trip upstairs?
3) The Girls, the Balls, and the Boxes
Four girls were blindfolded and each was given an identical box, containing different colored balls.
One box contained 3 black balls.
One box contained 2 black balls and 1 white ball.
One box contained 1 black ball and 2 white balls.
One box contained 3 white balls.
Each box had a label on it reading "BBB" (Three Black) or "BBW" (Two Black,
One White) or "BWW" (One Black, Two White) or "WWW" (Three White).
The girls were told that none of the four labels correctly described the contents of the box to which it was attached.
Each girl was told to draw two balls from her box, at which point the blindfold would be removed so that she could see the two balls in her hand and the label on the box assigned to her. She was given the task of trying to guess the color of the ball remaining in her box.
As each girl drew balls from her box, her colors were announced for all the girls to hear but the girls could not see the labels on any boxes other than their own.
The first girl, having drawn two black balls, looked at her label and announced: "I know the color of the third ball!"
The second girl drew one white and one black ball, looked at her label and similarly stated: "I too know the color of the third ball!"
The third girl withdrew two white balls, looked at her label, and said: "I can't tell the color of the third ball."
Finally, the fourth girl declared: "I don't need to remove my blindfold or any balls from my box, and yet I know the color of all three of them. What's more, I know the color of the third ball in each of the other boxes, as well as the labels of each of the boxes that you have."
The first three girls were amazed by the fourth girl's assertion and promptly challenged her. She proceeded to identify everything that she said she could.
Can you do the same?
Regards
Alan
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The Three Suspects
Kyle, Neal, and Grant were rounded up by their mother yesterday, because one of them was suspected of having grabbed a few too many cookies from the cookie jar. The three brothers made the following statements under very intensive questioning:
• Kyle: I'm innocent.
• Neal: I'm innocent.
• Grant: Neal is the guilty one.
Okay...
If Kyle is telling the truth, then Neal and Grant are lying - so Neal isn't innocent, but he is also not guilty.
If Neal is telling the truth, then Kyle and Grant are lying - so Kyle is guilty and Neal is not guilty
If Grant is telling the truth, then Kyle and Neal are guilty - so Kyle isn't innocent, and neither is Neal.
So... Neal is telling the truth, Kyle nicked the cookies and Grant, the little snitch, didn't do anything.
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A slightly quicker way to establishing Neal's innocence, without working out all the possibilities is that because Kyle's and Neal's statements are identical, one of them must be telling a lie and the other must be telling the truth. This means Grant's statement must be untrue and therefore identifies which one is telling the lie.
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Greetings
Benv,
Nice correct answer Ben.
1) The Three Suspects
Assume Neal took the cookies. If so, then both Kyle's and Grant's statements would be true. Hence it was not Neal.
Now assume Grant took the cookies. If so, then both Kyle's and Neal's statements would be true. Hence it was not Grant.
Now assume it was Kyle. This works...
"only Neal's statement was true"
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Three switches. If the lamps are of a type which gets warm...............
One on.
One off.
One on for a minute or so.
Could even do 4.. One on for several miniutes.. One on for 30 secs..
Bit like when detectives want to know if a car has been driven very recently.. The engine and exhaust will be hot.
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Pumblechook
You answer
Three switches. If the lamps are of a type which gets warm...............
One on.
One off.
One on for a minute or so.
Could even do 4.. One on for several miniutes.. One on for 30 secs..
Bit like when detectives want to know if a car has been driven very recently.. The engine and exhaust will be hot.
Good solution correct
Solved in much the same way!
The Three Switches
Turn Switch 1 on and leave it on for a little while... about five minutes or more... and then turn it off.
Turn Switch 2 on and go upstairs to inspect the lamps.
The lamp with the bulb that is off but warm is controlled by Switch 1.
The lamp that is currently on is controlled by Switch 2.
The lamp that is off and cold is controlled by Switch 3
Regards
Alan
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This from a free net site but wont give the link yet (no copyright)
“ Three gods A, B, and C are called, in some order, True, False, and Random. True always speaks truly, False always speaks falsely, but whether Random speaks truly or falsely is a completely random matter.
Your task is to determine the identities of A, B, and C by asking three yes-no questions; each question must be put to exactly one god.
The gods understand English, but will answer all questions in their own language, in which the words for yes and no are 'da' and 'ja', in some order. You do not know which word means which. ”
These the the clarifications
“ It could be that some god gets asked more than one question (and hence that some god is not asked any question at all).
What the second question is, and to which god it is put, may depend on the answer to the first question. (And of course similarly for the third question.)
Whether Random speaks truly or not should be thought of as depending on the flip of a coin hidden in his brain: if the coin comes down heads, he speaks truly; if tails, falsely.
Random will answer 'da' or 'ja' when asked any yes-no question.
Give it I go I will try without looking at the solution
Alan
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GUYS
AND GIRLS,
Here is a much easier puzzle I promise,
It doesn't hurt to take a hard look at yourself from time to time. This little test should help you get started.
During a visit to a mental asylum, a visitor asked the Director what the criteria is that defines if a patient should be institutionalized.
"Well," said the Director, "we fill up a bathtub. Then we offer a teaspoon, a teacup, and a bucket to the patient and ask the patient to empty the bathtub."
Okay, here's your test:
1. Would you use the spoon?
2. Would you use the teacup?
3. Would you use the bucket?
"Oh, I understand," said the visitor. "A normal person would choose the bucket, as it is larger than the spoon."
What was the director's response?
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Are you allowed to pull out the bathplug?
Did he lock up the visitor?
If he did, he would soon have his institution full of visitors.
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I can't help thinking that a normal person would tell the director to get lost.
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Or ask for some soap and a towel?
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sophiec,
Are you allowed to pull out the bathplug?
Did he lock up the visitor?
If he did, he would soon have his institution full of visitors
Yes Sophie silly little puzzle was it not
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Here is another easy one
A man is trapped in a room. The room has only two possible exits: two doors. Through the first door there is a room constructed from magnifying glass. The blazing hot sun instantly fries anything or anyone that enters. Through the second door there is a fire-breathing dragon.
How does the man escape?
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Another more based on astronomy
You awake inside a small transparent capsule sitting on the surface of Venus. From a small speaker you hear a voice that says, "We will leave you here either for a day or a year. If you choose to stay a day, we will give you $1 million.
If you choose to stay a year, we will give you $2 million.
Either way, you will have sufficient food and water. We will make sure the temperature is a constant 70 degrees Fahrenheit. We will also supply cable TV."
What is your choice? (Don't let money decide your answer).
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Here is another east one
A man is trapped in a room. The room has only two possible exits: two doors. Through the first door there is a room constructed from magnifying glass. The blazing hot sun instantly fries anything or anyone that enters. Through the second door there is a fire-breathing dragon.
How does the man escape?
He waits until night time and either just walks out through the now darkened magnifying glass room, or tip-toes past the sleeping dragon.
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Another more based on astronomy
You awake inside a small transparent capsule sitting on the surface of Venus. From a small speaker you hear a voice that says, "We will leave you here either for a day or a year. If you choose to stay a day, we will give you $1 million.
If you choose to stay a year, we will give you $2 million.
Either way, you will have sufficient food and water. We will make sure the temperature is a constant 70 degrees Fahrenheit. We will also supply cable TV."
What is your choice? (Don't let money decide your answer).
Is that a sidereal day or a solar day? I'd stay for the year, as it's shorter than the sidereal day (but longer than the solar day).
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LeeE
Both the anwers you gave are correct, on Venus the day is longer than the year.
The puzzle about the three gods is, however very difficult.
Alan
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Lightarrow,
Lightarrow,
I like you approach to this not so simple little problem
Hard Solution
It is intuitive that at all times the dogs will form the corners of a square, ever decreasing in size and rotating about the center.
Let C be the center of the square at point (0,0).
Let the dogs be at initial points (.5,.5), (.5,-.5), (-.5,-.5), and (-.5,.5).
Let dog 1 bet at (.5,-.5) who is chasing dog 2 at (.5,.5). Consider the line from C to dog 1.
Next consider the line from dog 1 to dog 2. The angle between these lines is 135 degrees.
Theorem: Let C be a fixed point, D a point on a curve, r the distance from C to D, x the polar angle formed by r, and y the angle formed by the angle between CD and the tangent line to the curve at point D. Then tan(y)=r/(dr/dx).
Can you find for me a prove of this theorem? I still haven't found on my books (but there must be somewhere).
So the line of sight from the center to the initial point of dog 1 would be be at a 270 degree angle. The line of sight from dog 1 to dog 2 would be a vertical line of 90 degrees. The angle between these lines is 270 degrees. The tangent of 270 is -1. So we have r/(dr/dx) = -1. Thus r(x)=c*e-x.
At the moment the dogs start walking x equals 0. At this moment the distance from the center to any of the dogs is 2-1/2. So r(x)=2-1/2*e-x.
The arc length is given by the formula:
Integral from 0 to infinity of ((f(x))2+(f'(x))2)1/2
And of this one also.
I only know this formula: L = ∫0+oo {[x'(t)]2 + [y'(t)]2}1/2dt.
Where x and y are the coordinates of the vector r(t) and t is the parameter of the curve (that is, the polar angle in this case, that you called "x"). =
Integral from 0 to infinity of ((2-1/2*e-x)2 + (-2-1/2*e-x)2)1/2 =
Integral from 0 to infinity of (2-1*e-2x + 2-1*e-2x)1/2 =
Integral from 0 to infinity of e-x =
-e-x from 0 to infinity =
=0 - (-e0) =
0 -(-1) = 1 The path the dogs take is called a logarithmic spiral.
It is an interesting paradox that the dogs will make an infinite number of circles, yet the total distance is constant. Every revolution the size of the square will to e-2*pi/21/2 = 0.03055663 times it's size before the revolution.
Easy Solution
David Wilson suggested the more simple solution which follows...
Suppose dog A is pursuing dog B who is pursuing dog C. During the entire pursuit, the dogs remain at the corners of a square, and angle ABC is a constant 90 degrees. That is, B's path towards C is always perpendicular to A's path towards B, and B has zero velocity in the direction along A's path. Since B neither approaches nor recedes from A during the walk, A simply covers the intial separation of 1.
This is very interesting, clever solution!
Antonio Molins went further to state that as dog A is heading direcctly towards dog B, dog B is moving towards from dog A at rate equal to cosine of any of the interior angles of the shape made by the dogs.
He also provided the following graphics.
For example consider a pentagon. Remember that for a n-gon of n sides each angle will be 180*(1-(2/n)).
So a pentagon will have angles of 108°. Let's assume each dog walks at a rate of 1 unit per minute, where the initial shape has sides of 1 unit. As pointed out by David Wilson we can forget about the spiral paths but instead think of the dogs walking in straight lines.
In the case of the penagon for every one unit dog A draw towards dog B's initial position, dog B will draw cos(180°)=-0.309
cos(180°) = -1units towards dog A.
The question is, how long will it take for them to meet? Let the answer to that question be t. In t seconds dog A will cover t units, and dog B will cover .309*t units. The total distance traveled by A equals the distance traveled by B plus 1 (the initial separation). So solving for t: t=0.309+1 --> t=1/(1-.309) --> t=1.4472.
What do you think?
T = a/v, as I've written in my previous post. If a = 1 and v = 1 unit/minute, then T = 1 minute.
Very interesting problem however, thank you!
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BoardChemist,
I don't know how to hide an answer, but I doubt this is the answer being looked for
.
"" to the mind guess"" of prediction test in "blue letters", This is post separate from the horse hay one.
The puzzle states that the hay is tied down. It does not state that the stake is driven into the ground so maybe the horse just walks over to the hay dragging the stake.(Though my first thought
was that he chews through, or breaks the rope.)
Yessssssss!! The above is the correct answer a little silly I know but many people most often just presume the skate is firmly hamered into the ground
Regards
Alan
What's wrong with LeeE's answer?:
The horse is standing at the extreme length of the rope, five meters from the stake. The horse is facing the stake, and five meters further away in the same direction is the hay. The horse walks the ten meters across the 'circle' and eats the hay
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I suppose it could have been a sawing horse - so no problem, again.
[:D]
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Two glasses, one nearly filled with liquid A, one with liquid B. Take a teaspoon of A and put it in B, stir and put a teaspoon of the mixture back in A. Which now is the purest??
None, if they initially contains the same volume of liquids of the same density (and the density doesn't vary with mixing).
I usually propose to people, at breakfast, a slight variation of this problem:
you take a teaspoon of coffe from a little cup and put it in a large cup of milk [so the initial amounts of coffee and milk can be very different]. Then you mix the milk-coffee, you take a teaspoon of it and put it in the little cup of coffee. Is there more coffee in the cup of milk or milk in the little cup of coffee?
Let be v the volume in the teaspoon, so you take away v of coffee from the little cup and put it in the large cup. Now you take a teaspoon of milk-coffee: this will contain x of coffee and v-x of milk, and you put it in the little cup, so it contains now v-x of milk. Now here you have taken away v of coffee and re-added x of it so you have taken away v-x of coffee from the little cup and so this is the amount of coffee you have put in the cup of milk, so the two amounts are the same.
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MathProblems.info
Problem 29 Solution
Hard Solution
It is intuitive that at all times the dogs will form the corners of a square, ever decreasing in size and rotating about the center.
Let C be the center of the square at point (0,0). Let the dogs be at initial points (.5,.5), (.5,-.5), (-.5,-.5), and (-.5,.5). Let dog 1 bet at (.5,-.5) who is chasing dog 2 at (.5,.5). Consider the line from C to dog 1. Next consider the line from dog 1 to dog 2. The angle between these lines is 135 degrees.
Theorem: Let C be a fixed point, D a point on a curve, r the distance from C to D, x the polar angle formed by r, and y the angle formed by the angle between CD and the tangent line to the curve at point D. Then tan(y)=r/(dr/dx).
So the line of sight from the center to the initial point of dog 1 would be be at a 270 degree angle. The line of sight from dog 1 to dog 2 would be a vertical line of 90 degrees. The angle between these lines is 270 degrees. The tangent of 270 is -1. So we have r/(dr/dx) = -1. Thus r(x)=c*e-x.
At the moment the dogs start walking x equals 0. At this moment the distance from the center to any of the dogs is 2-1/2. So r(x)=2-1/2*e-x.
The arc length is given by the formula:
Integral from 0 to infinity of ((f(x))2+(f'(x))2)1/2 =
Integral from 0 to infinity of ((2-1/2*e-x)2 + (-2-1/2*e-x)2)1/2 =
Integral from 0 to infinity of (2-1*e-2x + 2-1*e-2x)1/2 =
Integral from 0 to infinity of e-x =
-e-x from 0 to infinity =
=0 - (-e0) =
0 -(-1) = 1 The path the dogs take is called a logarithmic spiral. It is an interesting paradox that the dogs will make an infinite number of circles, yet the total distance is constant. Every revolution the size of the square will to e-2*pi/21/2 = 0.03055663 times it's size before the revolution.
--------------------------------------------------------------------------------
Easy Solution
David Wilson suggested the more simple solution which follows...
Suppose dog A is pursuing dog B who is pursuing dog C. During the entire pursuit, the dogs remain at the corners of a square, and angle ABC is a constant 90 degrees. That is, B's path towards C is always perpendicular to A's path towards B, and B has zero velocity in the direction along A's path. Since B neither approaches nor recedes from A during the walk, A simply covers the intial separation of 1.
Antonio Molins went further to state that as dog A is heading direcctly towards dog B, dog B is moving towards from dog A at rate equal to cosine of any of the interior angles of the shape made by the dogs. He also provided the following graphics.
For example consider a pentagon. Remember that for a n-gon of n sides each angle will be 180*(1-(2/n)). So a pentagon will have angles of 108°. Let's assume each dog walks at a rate of 1 unit per minute, where the initial shape has sides of 1 unit. As pointed out by David Wilson we can forget about the spiral paths but instead think of the dogs walking in straight lines. In the case of the penagon for every one unit dog A draw towards dog B's initial position, dog B will draw cos(180°)=-0.309 units towards dog A. The question is, how long will it take for them to meet? Let the answer to that question be t. In t seconds dog A will cover t units, and dog B will cover .309*t units. The total distance traveled by A equals the distance traveled by B plus 1 (the initial separation). So solving for t: t=0.309+1 --> t=1/(1-.309) --> t=1.4472.
Thus the general formula is 1/(1+cos(t)), where t is an interior angle of the shape.
Number of sides = 3
Lengh of initial side = 1
Each angle = 180*(1-(2/3)) = 60°
Length of path = 1/(1+cos(60)) = 1/(1+0.5) = 2/3
Number of sides = 4
Lengh of initial side = 1
Each angle = 180*(1-(2/4)) = 90°
Length of path = 1/(1+cos(90)) = 1/(1+0) = 1
Number of sides = 5
Lengh of initial side = 1
Each angle = 180*(1-(2/5)) = 108°
Length of path = 1/(1+cos(108)) = 1/(1-.309) = 1.4472
Number of sides = 6
Lengh of initial side = 1
Each angle = 180*(1-(2/6)) = 120°
Length of path = 1/(1+cos(120)) = 1/(1-.5) = 2
Number of sides = 10
Lengh of initial side = 1
Each angle = 180*(1-(2/10)) = 144°
Length of path = 1/(1+cos(144)) = 1/(1-.809) = 5.2361
Number of sides = 15
Lengh of initial side = 1
Each angle = 180*(1-(2/15)) = 156°
Length of path = 1/(1+cos(156)) = 1/(1-.9135) = 11.5668
Michael Shackleford, A.S.A.
MathProblems.info home
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LightArrow
I am impressed by your solutions to these difficult problems!
I could give you the link by PM, your maths is better than mine.
perhaps you could then post some math problems from this site
Regards
Alan
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LightArrow
I am impressed by your solutions to these difficult problems!
I could give you the link by PM, your maths is better than mine.
perhaps you could then post some math problems from this site
Regards
Alan
Yes, I'm interested, thank you.
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Easy Solution
David Wilson suggested the more simple solution which follows...
Suppose dog A is pursuing dog B who is pursuing dog C. During the entire pursuit, the dogs remain at the corners of a square, and angle ABC is a constant 90 degrees. That is, B's path towards C is always perpendicular to A's path towards B, and B has zero velocity in the direction along A's path. Since B neither approaches nor recedes from A during the walk, A simply covers the intial separation of 1.
It cannot be as simple as that; if I am chasing someone and, eventually, I catch them, you can't say that the distance I have run is just our initial separation; I may have to run miles and miles before catching them.
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Sophie correct it is not as simple as that, read light arrows postshe made a valiant effort.
The easy solution just gives an approximation
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It does give a sort of lower limit, I grant you, but that's about all.
The shortest way for them to get together (the actual lower limit) is if they all aim at the centre - then they travel Root 2 of their initial separation.
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Do you know this problem:
There is a solid sphere (full inside) and you make a hole in it with a drill which go straight through the centre of the sphere to the opposite side (you probably have in english a better way to describe such a kind of hole).
The hole is 6 cm long.
Which is the remaining (residual) volume of the sphere?
"But, wait a moment, you didn't tell us the sphere's radius!" Exactly!
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Are you sure there's nothing else you need to tell us?
Why couldn't you drill a 6cm diameter hole through the Earth or through a football?
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I think this is the problem lightarrow's thinking of:
Old Boniface he took his cheer,
Then he bored a hole through a solid sphere,
Clear through the center, straight and strong,
And the hole was just six inches long.
Now tell me, when the end was gained,
What volume in the sphere remained?
Sounds like I haven't told enough,
But I have, and the answer isn't tough!
This can be worked out mathematically or logically (I solved it via logic).
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Are you sure there's nothing else you need to tell us?
Why couldn't you drill a 6cm diameter hole through the Earth or through a football?
Sorry, it's 6 cm long, not in diametre. I've correct the post.
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I think this is the problem lightarrow's thinking of:
Old Boniface he took his cheer,
Then he bored a hole through a solid sphere,
Clear through the center, straight and strong,
And the hole was just six inches long.
Now tell me, when the end was gained,
What volume in the sphere remained?
Sounds like I haven't told enough,
But I have, and the answer isn't tough!
This can be worked out mathematically or logically (I solved it via logic).
Exactly that problem, right.
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So we can assume the answer is always the same so an infinitely thin drill through a 6cm diameter sphere would leave 4XPiX33/4 material behind - i.e. taking nothing out of the sphere.
Proving that it works for a range of sphere sizes is a bit harder - I will try it sometime. The question implies that original sphere could be ANY radius greater than 3cm. Wierd.
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So we can assume the answer is always the same so an infinitely thin drill through a 6cm diameter sphere would leave 4XPiX33/4 material behind - i.e. taking nothing out of the sphere.
Proving that it works for a range of sphere sizes is a bit harder - I will try it sometime. The question implies that original sphere could be ANY radius greater than 3cm. Wierd.
(4/3)π33 = 36π.
Apart the mistake in the formula for a sphere's volume ( [:)]) you answered correctly. Clearly with more computations you can show that the remaining volume doesn't depend on the radius (if ≥ 3). Weird, yes.
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I was so pre-occupied with putting in the supercripts that I missed that!
Durrr!
It could work, approximately, for olives too!
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Hold on, you guys have confused me...
If you were to drill a hole through a sphere, right through the centre, and the hole was 6cm long, then the sphere must have a diameter of 6cm , no?
In which case, surely we need to know the diameter of the bore hole to determine how much volume is lost - i.e. volume of (6cm diameter) sphere, minus volume of (6cm length, n diameter) cylinder?
Edit - I'm basing this on the idea that drilling through something involves coming out of the other side - drilling 6cm into something is different.
Edit2 - And going back to the horse thing - if 'the stake wasn't hammered into the ground' is a valid answer, then so is 'he chewed through the rope', 'the stake was between him and the hay' and 'a kindly old man was passing and moved the hay closer to him'.
Edit3 - I've just realised that I may be a pedant...
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Hold on, you guys have confused me...
If you were to drill a hole through a sphere, right through the centre, and the hole was 6cm long, then the sphere must have a diameter of 6cm , no?
Make a draw of the bored sphere seen from a lateral view. How do you measure the hole's lenght? [ Invalid Attachment ]
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The question specifically doesn't mention the diameter of the hole, so assuming that it's not a trick question, which it isn't, the significance of not specifying the hole diameter is that the answer is the same, regardless of the diameter of the hole.
Thus, sophiecentaur's approach is the correct logical solution.
As long as the length of the hole remains the same, any diameter will give the same answer.
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Guys,
I decided to post the solution,
The Hardest Logic Puzzle Ever
“ Three gods A, B, and C are called, in some order, True, False, and Random. True always speaks truly, False always speaks falsely, but whether Random speaks truly or falsely is a completely random matter. Your task is to determine the identities of A, B, and C by asking three yes-no questions; each question must be put to exactly one god. The gods understand English, but will answer all questions in their own language, in which the words for yes and no are 'da' and 'ja', in some order. You do not know which word means which. ”
Boolos provides the following clarifications:[1]
“ It could be that some god gets asked more than one question (and hence that some god is not asked any question at all).
What the second question is, and to which god it is put, may depend on the answer to the first question. (And of course similarly for the third question.)
Whether Random speaks truly or not should be thought of as depending on the flip of a coin hidden in his brain: if the coin comes down heads, he speaks truly; if tails, falsely.
Random will answer 'da' or 'ja' when asked any yes-no question.[1]
”
Hence whenever you ask them a yes-no question, they reply 'Bal' or 'Da' - one of which means yes and the other no. The trouble is that we do not know which of 'Bal' or 'Da' means yes and which means no". There are other related puzzles in The Riddle of Scheherazade (see especially, p. 114).
More generally this puzzle is based on Smullyan's famous Knights and Knaves puzzles (e.g. on a fictional island, all inhabitants are either knights, who always tell the truth, or knaves, who always lie. The puzzles involve a visitor to the island who must ask a number of yes/no questions in order to discover what he needs to know).
A version of these puzzles was popularized by a scene in the 1980's fantasy film, Labyrinth. There are two doors with two guards. One guard lies and one guard doesn't. One door leads to the castle and the other leads to "certain death". The puzzle is to find out which door leads to the castle by asking one of the guards one question. In the movie Sarah does this by asking "Would he tell me that this door leads to the castle?".
[edit] The solution
Boolos provided his solution in the same article in which he introduced the puzzle. Boolos states that the "first move is to find a god that you can be certain is not Random, and hence is either True or False".[1] There are many different questions that will achieve this result. One strategy is to use complicated logical connectives in your questions (either biconditionals or some equivalent construction).
Boolos' question was:
Does 'da' mean yes if and only if you are True if and only if B is Random?[1]
Equivalently:
Are an odd number of the following statements true: you are False, 'ja' means yes, B is Random?
The puzzle's solution can be simplified by using counterfactuals.[2][3] The key to this solution is that, for any yes/no question Q, asking either True or False the question
If I asked you Q, would you say 'ja'?
results in the answer 'ja' if the truthful answer to Q is yes, and the answer 'da' if the truthful answer to Q is no. The reason this works can be seen by looking at the eight possible cases.
Assume that 'ja' means yes and 'da' means no.
(i) True is asked and responds with 'ja'. Since he is telling the truth the truthful answer to Q is 'ja', which means yes.
(ii) True is asked and responds with 'da'. Since he is telling the truth the truthful answer to Q is 'da', which means no.
(iii) False is asked and responds with 'ja'. Since he is lying it follows that if you asked him Q he would instead answer 'da'. He would be lying, so the truthful answer to Q is 'ja', which means yes.
(iv) False is asked and responds with 'da'. Since he is lying it follows that if you asked him Q he would in fact answer 'ja'. He would be lying, so the truthful answer to Q is 'da', which means no.
Assume 'ja' means no and 'da' means yes.
(v) True is asked and responds with 'ja'. Since he is telling the truth the truthful answer to Q is 'da', which means yes.
(vi) True is asked and responds with 'da'. Since he is telling the truth the truthful answer to Q is 'ja', which means no.
(vii) False is asked and responds with 'ja'. Since he is lying it follows that if you asked him Q he would in fact answer 'ja'. He would be lying, so the truthful answer to Q 'da', which means yes.
(viii) False is asked and responds with 'da'. Since he is lying it follows that if you asked him Q he would instead answer 'da'. He would be lying, so the truthful answer to Q is 'ja', which means no.
Using this fact, one may proceed as follows.[2]
Ask god B, "If I asked you 'Is A Random?', would you say 'ja'?". If B answers 'ja', then either B is Random (and is answering randomly), or B is not Random and the answer indicates that A is indeed Random. Either way, C is not Random. If B answers 'da', then either B is Random (and is answering randomly), or B is not Random and the answer indicates that A is not Random. Either way, A is not Random.
Go to the god who was identified as not being Random by the previous question (either A or C), and ask him: "If I asked you 'Are you True?', would you say 'ja'?". Since he is not Random, an answer of 'ja' indicates that he is True and an answer of 'da' indicates that he is False.
Ask the same god the question: "If I asked you 'Is B Random?', would you say 'ja'?". If the answer is 'ja' then B is Random; if the answer is 'da' then the god you have not yet spoken to is Random. The remaining god can be identified by elimination.
[edit] Random's behaviour
Most readers of the puzzle assume that Random will provide completely random answers to any question asked of him; however, the puzzle does not actually state this. In fact, Boolos' third clarifying remark explicitly refutes this assumption.[2]
Whether Random speaks truly or not should be thought of as depending on the flip of a coin hidden in his brain: if the coin comes down heads, he speaks truly; if tails, falsely.
This says that Random randomly acts as a liar or a truth-teller, not that Random answers randomly.
A small change to the question above yields a question which will always elicit a meaningful answer from Random. The change is as follows:
If I asked you Q in your current mental state, would you say 'ja'?[2]
We have effectively extracted the truth-teller and liar personalities from Random and forced him to be only one of them. This completely trivializes the puzzle since we can now get truthful answers to any questions we please.
1. Ask god A, "If I asked you 'Are you Random?' in your current mental state, would you say 'ja'?"
If A answers 'ja', then A is Random:
2a. Ask god B, "If I asked you 'Are you True?', would you say 'ja'?"
If B answers 'ja', then B is True and C is False.
If B answers 'da', then B is False and C is True. In both cases, the puzzle is solved.
If A answers 'da', then A is not Random:
2b. Ask god A, "If I asked you 'Are you True?', would you say 'ja'?"
If A answers 'ja', then A is True.
If A answers 'da', then A is False.
3. Ask god A, "If I asked you 'Is B Random?', would you say 'ja'?"
If A answers 'ja', then B is Random, and C is the opposite of A.
If A answers 'da', then C is Random, and B is the opposite of A.
We can modify Boolos' puzzle so that Random is actually random by replacing Boolos' third clarifying remark with the following.
Whether Random says 'ja' or 'da' should be thought of as depending on the flip of a coin hidden in his brain: if the coin comes down heads, he says 'ja'; if tails, he says 'da'.
With this modification, the puzzle's solution demands the more careful god-interrogation given at the end of the The Solution section.
[edit] Exploding god-heads
In A simple solution to the hardest logic puzzle ever,[2] B. Rabern and L. Rabern develop the puzzle further by pointing out that it is not the case that 'ja' and 'da' are the only possible answers a god can give. It is also possible for a god to be unable to answer at all. For example, if the question "Are you going to answer this question with the word that means no in your language?" is put to True, he cannot answer truthfully. (The paper represents this as his head exploding, "...they are infallible gods! They have but one recourse – their heads explode") Allowing the "exploding head" case gives yet another solution of the modified puzzle (modified so that Random is actually random) and introduces the possibility of solving the original puzzle (unmodified) in just two questions rather than three. In support of a two-question solution to the puzzle, the authors solve a similar simpler puzzle using just two questions.
Three gods A, B, and C are called, in some order, Zephyr, Eurus, and Aeolus. The gods always speak truly. Your task is to determine the identities of A, B, and C by asking three yes-no questions; each question must be put to exactly one god. The gods understand English and will answer in English.
Note that this puzzle is trivially solved with three questions (just ask away!). To solve the puzzle in two questions, the following lemma is proved.
Tempered Liar Lemma. If we ask A "Is it the case that {[(you are going to answer 'no' to this question) AND (B is Zephyr)] OR (B is Eurus) }?", a response of 'yes' indicates that B is Eurus, a response of 'no' indicates that B is Aeolus, and an exploding head indicates that B is Zephyr. Hence we can determine the identity of B in one question.
Using this lemma it is simple to solve the puzzle in two questions. A similar trick (tempering the liar's paradox) can be used to solve the original puzzle in two questions.
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I decided to post the solution,
...
The Hardest Logic Puzzle Ever
...
GULP!
Ehm...
Something a bit more simple:
A sample of 100 kg of potatoes has a water content of 99%. It is put under the sun to dry a little, and the water content at the end becomes 98%. How does the potatoes weight now?
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A sample of 100 kg of potatoes has a water content of 99%. It is put under the sun to dry a little, and the water content at the end becomes 98%. How does the potatoes weight now?
I know it's simple, but the result is quite amazing.
Anyway, if you don't like that problem, try this:
you have two fuses which burn exactly in one hour each; they don't burn in a regular way, however, so you can't say that half lenght will burn in half an hour, and so on.
The problem is that you have to measure exactly 45 minutes. How will you do it?
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[quote author=Alan McDougall link=topic=15562.msg188243#msg188243 date=1217501149
Ehm...
A sample of 100 kg of potatoes has a water content of 99%. It is put under the sun to dry a little, and the water content at the end becomes 98%. How does the potatoes weight now?
Is the water content 97.02%? Do the potatoes weigh 98kgs?
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A sample of 100 kg of potatoes has a water content of 99%. It is put under the sun to dry a little, and the water content at the end becomes 98%. How does the potatoes weight now?
Is the water content 97.02%? Do the potatoes weigh 98kgs?
No, the water content, at the end, is exactly 98% and the potatoes weight is not 98 kg.
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you have two fuses which burn exactly in one hour each; they don't burn in a regular way, however, so you can't say that half lenght will burn in half an hour, and so on.
The problem is that you have to measure exactly 45 minutes. How will you do it?
My solution requires two conditions:-
a) that both fuses, although they do not burn at a constant rate, are identical: their varying burn rate is the same in both fuses.
b) it is possible to discriminate the ends of the fuses, so they can both be lit from the same end.
Solution:-
1) place one fuse in a loop, to form the letter “O”, so both ends of the fuse can be lit simultaneously.
When the two flames meet in the “O” half an hour will have elapsed. Mark the point on the fuse at which the two flames meet,
(this point is probably not the midpoint as the fuse did not burn at a uniform rate).
2) Transfer this mark to the same position on the second fuse. Make this fuse into a letter “P” with the loop of the “P” starting at this mark. Light the base of the “P”, when the single flame reaches the loop of the “P” half an hour will have elapsed, the loop of the “P” will then burn with two flames like the “O”, the loop of the “P“ will burn in 15 minutes. So the “P” fuse will take 45 minutes to burn out.
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It's possible to solve it without your assumption a) but with b) only.
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Hello,
Here are a few more problems to solve, the first few easy and the last a little more difficult. If interest is shown I will come in with other perplexing puzzles or problems of logic some more difficult and the others much easier to keep the topic as fun thread
1) The Frog
A frog is at the bottom of a 30 meter well. Each day he summons enough energy for one 3 meter leap up the well. Exhausted, he then hangs there for the rest of the day. At night, while he is asleep, he slips 2 meters backwards. How many days does it take him to escape from the well?
Note: Assume after the first leap that his hind legs are exactly three meters up the well. His hind legs must clear the well for him to escape.
Answer??
2) The Socks
Cathy has six pairs of black socks and six pairs of white socks in her drawer.
In complete darkness, and without looking, how many socks must she take from the drawer in order to be sure to get a pair that match?
Answer??
3) There is something about Mary
Mary's mum has four children.
The first child is called April.
The second May.
The third June.
What is the name of the fourth child?
Answer ??
4) 100 Gold Coins (this problem is more difficult)
Five pirates have obtained 100 gold coins and have to divide up the loot. The pirates are all extremely intelligent, treacherous and selfish (especially the captain).
The captain always proposes a distribution of the loot. All pirates vote on the proposal, and if half the crew or more go "Aye", the loot is divided as proposed, as no pirate would be willing to take on the captain without superior force on their side.
If the captain fails to obtain support of at least half his crew (which includes himself), he faces a mutiny, and all pirates will turn against him and make him walk the plank. The pirates start over again with the next senior pirate as captain.
What is the maximum number of coins the captain can keep without risking his life?
Answer?
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It's possible to solve it without your assumption a) but with b) only.
Ok, I'll give the solution, it seems a very difficult problem!
1.Light simultaneously the two ends of the first fuse and one end of the second.
2.After exactly 30 minutes the first fuse will be completely burnt and the second will have exactly other 30 minutes to burn.
3.At this moment, light the second end of the second fuse.
4.After exactly 15 minutes from that moment, the second fuse will be completely burnt. 30 + 15 = 45 minutes.
I have to admit that, even if the solution is quite simple, I wasn't able to find it.
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1) The Frog
A frog is at the bottom of a 30 meter well. Each day he summons enough energy for one 3 meter leap up the well. Exhausted, he then hangs there for the rest of the day. At night, while he is asleep, he slips 2 meters backwards. How many days does it take him to escape from the well?
Note: Assume after the first leap that his hind legs are exactly three meters up the well. His hind legs must clear the well for him to escape.
Answer??
29 days, I suppose!
2) The Socks
Cathy has six pairs of black socks and six pairs of white socks in her drawer.
In complete darkness, and without looking, how many socks must she take from the drawer in order to be sure to get a pair that match?
Answer??
3 socks, I suppose!
3) There is something about Mary
Mary's mum has four children.
The first child is called April.
The second May.
The third June.
What is the name of the fourth child?
Answer ??
Mary, I suppose!
4) 100 Gold Coins (this problem is more difficult)
Five pirates have obtained 100 gold coins and have to divide up the loot. The pirates are all extremely intelligent, treacherous and selfish (especially the captain).
The captain always proposes a distribution of the loot. All pirates vote on the proposal, and if half the crew or more go "Aye", the loot is divided as proposed, as no pirate would be willing to take on the captain without superior force on their side.
If the captain fails to obtain support of at least half his crew (which includes himself), he faces a mutiny, and all pirates will turn against him and make him walk the plank. The pirates start over again with the next senior pirate as captain.
What is the maximum number of coins the captain can keep without risking his life?
No idea how to solve it...
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Me too - I can solve it in two weighings if I cheat and use either gravitational attraction or momentum, but not purely via weighing.
can u be more explicit with those methods please.
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He could propose to take 32 coins for himself and offer 34 each to 2 members of his crew. Content enough that no one will receive any more than them, those 2 should vote in his favour. The captain himself will vote in favour as well, giving a total of 3 votes, enough to save his life.
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On second thoughts, take 33 coins and offer 33 to 2 others, and 1 coin to split between the other 2. Same result.
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The problem is that the two crew members he would hope to support him could probably do better by rejecting the offer.
I suggest that there would be no objection from his two 'supporters' if he offered them a lot more.
If he offered them 33 each and kept 34 for himself, they couldn't expect more than that so they would be bound to support him.
BUT there is probably a smarter answer which would give him more in the end.
If they go to the 'next round' after a mutiny, the new boss would still need to pay out to two others for a majority and he might not choose one of the original two. So they would possibly settle for a smaller cut. But, how to work it out???
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FROG - 28 days. After 27 days, he has reached 27 metres. On the 28th day, he can jump the final 3 metres and reach the top. This assumes that his first jump is on day 1, not day 0.
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Socks: 3.
The options:
A) pick 1: B pick 2: B ==> B pair
B) pick 1: B pick 2: W pick 3: B ==> B pair
C) pick 1: B pick 2: W pick 3: W ==> W pair
D) pick 1: W pick 2: W ==> W pair
E) pick 1: W pick 2: B pick 3: B ==> B pair
E) pick 1: W pick 2: B pick 3: W ==> W pair
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4th Child: Mary
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pirates:
Could it be that the captain can retain 98 coins, with 2 other pirates each having 1 making 100. The other 2 pirates who don't get a coin won't agree, but since they are in the minority, they won't mutiny.
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A sample of 100 kg of potatoes has a water content of 99%. It is put under the sun to dry a little, and the water content at the end becomes 98%. How does the potatoes weight now?
So 100 kg pots = 99% water plus 1% "stuff".
Therefore 1% stuff = 1kg and 99% water = 99kg.
98% water = 98 kg.
Pots = 1kg "stuff" plus 98kg = 99kg.
???
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Can the captain rely on the two accepting the one coin each? All four crew could vote against him. Deal or No Deal demonstrates that people will reject a small prize if a larger one is still available- even if there is a chance of no prize.
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A sample of 100 kg of potatoes has a water content of 99%. It is put under the sun to dry a little, and the water content at the end becomes 98%. How does the potatoes weight now?
So 100 kg pots = 99% water plus 1% "stuff".
Therefore 1% stuff = 1kg and 99% water = 99kg.
98% water = 98 kg.
Pots = 1kg "stuff" plus 98kg = 99kg.
???
No.
Hint: "to dry a little" means that the water content goes from 99% to 98%, not that the weight they lose is little...
When you will know the solution you will be surprised.
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OK..99% water means 1% "stuff".
Now, afterwards we have 2% "stuff".
OK. if 2%=1kg then the total = 1/0.02 = 50kg.
The answer is therefore 50kg.