That is clearly a much more elegant solution that my own. I concede my turn to you, good sir. ;) :DQuote:
Originally Posted by Jimbo
really, if someone wants to riddle, go ahead. I've always much preferred answering riddles than posing them.
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That is clearly a much more elegant solution that my own. I concede my turn to you, good sir. ;) :DQuote:
Originally Posted by Jimbo
really, if someone wants to riddle, go ahead. I've always much preferred answering riddles than posing them.
You are on an island.
On the island is a cave.
In the cave is an idol. The idol is either black, or it is white. Your life depends on knowing the correct color, but you may not enter the cave or see the idol.
There are two separate tribes on the island.
One tribe always tells the truth...
The other always lies.
You can not know which tribe is truth tellers, or which is liars.
Both tribes know the true color of the idol.
You may ask only one question, one time, to one tribe, to determine the correct color of the idol.
What is the one question that will allow you to correctly determine the true color of the idol?
Scott
Whoops!
James.
Ask: "If I asked the other tribe what colour the idol is, what would they say?"Quote:
Originally Posted by honedright
Whatever answer you are given the opposite is true.
(I learned this on an old episode of Doctor Who. :D)
I learned it watching the labyrinth!Quote:
Originally Posted by Rajagra
I want to know the answer to the elevator question now, why not repost it? :)
Why couldn't youu just ask a tribe to open the door to the cave?
James.
A man lives on the 30th floor of an apartment building.
Each day he takes the elevator from his flat to the ground floor. However, his return trip on the elevator is slightly more complicated: On sunny days he take the elevator to the 20th floor and walks up the steps to his flat on the 30th. On rainy days he takes the elevator all the way to the 30th floor.
Why?
James.
Correct.Quote:
Originally Posted by Rajagra
I first heard this riddle back in the 70's. Never watched Dr. Who so didn't know it was "out there."
Too easy, ah well.
Scott
On rainy days he's wearing rain boots and can reach the 30th floor button? He's very short.Quote:
Originally Posted by Jimbo
Scott
Because that's how the riddle is constructedQuote:
Originally Posted by Jimbo
I suspect that's close enough. Maybe a short person with an umbrella on rainy days?Quote:
Originally Posted by honedright
By the way, read this. :)
Yes - he is a dwarf and on rainy days he has his umbrella with him. So he can reach the higher buttons in the elevator with his umbrella on rainy days.
I think Scott was pretty much on the money.
James.
Hello gents...
One quick riddle for you...
What does a blind man see, a deaf man hear, and if you eat it, you are sure to die?
Nothing. This message is too short. So there.
Why wouldn't this guy just always carry a stick (or umbrella?) Sorry....Quote:
Originally Posted by Jimbo
Well, I guess he could do that.... Maybe he is a small person with a deep-seeded sense of umbrella propriety.Quote:
Originally Posted by smokelaw1
James.
OK. Here's one that should get you going..
If n is an integer greater than two prove, without recourse to the Taniyama-Shimura Conjecture, that the equation a^n + b^n = c^n has no solutions for non-zero integers a, b, and c.
Note: you may write your solutions in the margin if you wish. If the margin should prove too narrow to contain your proof, you may leave your solution tantalisingly unwritten for several centuries.
James.
OK, I'm done - since we're 21st century I typed it in the margin of my browser. It fit because I used a font size that ensures it.
If you want to see it you'll have to wait for the next generation of web browsers where you could see my browser as well, not just yours. I don't think it'll take you few centuries, but who knows... I don't make the technology.
Does gugi's answer satisfy the riddle-writer?
Since I trust Gugi implicitly, yes, I am satisfied with his answer.
Next riddle please!!
James.
@$$#ø|€ ;)
calligraphy?Quote:
Originally Posted by xman
I shall warn you that this one may be a bit challenging - it took me about 20min the first time I heard it, but I like a good challenge. If it fails I'll come up with another one
you have 13 balls and one of them is different weigh than the rest.
with only 3 measurements using simple scales, i.e. equal weight or not between groups of balls, find the odd ball.
And I guess this confession also makes your trust explicit from now on :)Quote:
Originally Posted by Jimbo
Are you told which side is heaviest, or just given binary info i.e. same vs not the same?Quote:
Originally Posted by gugi
Also, can you "cheat" by observing what happens as you remove one ball from each side after each weighing?
OK I can prove it. Next question?Quote:
Originally Posted by Jimbo
Lets call the 1-13. Ignore the thirteenth one for now.
First Usage:
Weigh 4 weights on each side (remember which ones and the slope):
eg. 1-2-3-4 vs 5-6-7-8
Unbalance --> < A > (answer in one of these 8 weights)
Balance --> < B > (answer in other 4 weights)
Second Usage
< A >
Of the 8 weights you weighed before, take 3 off the right side, and
grab 2 from the left side and put it them on the right. And then
one weight you didn't weigh before on the left.
eg. 1-2-x vs 3-4-5
(where x=9, 10, 11 or 12)
Slope of Balance Unchanged --> <A-1>
Slope of Balance Changed --> <A-2>
Balance --> <A-3>
< B >
Leave 3 weights on one side, and take 3 weights you didn't weigh
before and put them one the other side.
eg. 1-2-3 vs 9-10-11
Right Heavier --> <B-1>
Left Heavier --> <B-2>
Balance --> ANSWER = Weight never weighed
Third Usage
<A-1>
Since the slope is unchanged, the two you switched from left to
right before and are not the answer. The answer is one of the other
three weights, the other 9 are "wrong". Now, move the "answer"
weight from the right and put it on the left, and take the other
"answer" weight (on the left) out. Now put two of the other "wrong"
weights on the right side.
eg. 1-5 vs x-x
(where x=3, 4, 6, 7, 8, 9, 10, 11 or 12)
Slope of Balance Changed --> ANSWER = Weight on right in 2nd
usage
Slope of Balance Unchanged --> ANSWER = Weight on left in 2nd
usage
Balance --> ANSWER = Weight left out
<A-2>
Since the slope has changed, the one of two weights you switched
from left to right before are the answer. Move one of the two back
to the left side.
eg. 3 vs 4
Slope of Balance Changed --> ANSWER = Left weight
Slope of Balance Unchanged --> ANSWER = Right Weight
<A-3>
Since the scale is balanced, the answer is one of the 3 weights you
had taken out. Put one on each side of the scale and leave the
other out.
eg. 6 vs 7
Unbalance --> If slope in First Usage said that right side was
heavier, ANSWER = Heavier weight
--> If slope in First Usage said that right side was
lighter, ANSWER = Lighter weight
Balance --> ANSWER = Weight left out
<B-1>
The three on the right are heavier, so put one on each side of the
scale, and leave one out.
eg. 9 vs 10
Unbalance --> ANSWER = Heavier side
Balance --> ANSWER = Weight left out
<B-2>
The three on the right are lighter, so put one on each side of the
scale, and leave one out.
eg. 9 vs 10
Unbalance --> ANSWER = Lighter side
Balance --> ANSWER = Weight left out</pre>IF all measurements are always equal, then the thirteenth ball is of a different weight.
Clear as mud?
yes you would know which side is heavier.Quote:
Originally Posted by Rajagra
I don't think this helps you at all - presume that the difference in weight between the balls is much smaller than their individual weight. But if cheating helps you I'd be interested to know how.Quote:
Also, can you "cheat" by observing what happens as you remove one ball from each side after each weighing?
Yes Sandy, looks like you're right.Quote:
Originally Posted by sidneykidney
My wording (which may be clearer or not) would be:
First split them in three groups A: 4 balls, B:4 balls, C: 5balls
Weighing A vs B you'd narrow it to either the last 5 balls (A=B) or to the first 8 (A!=B).
I.) In case it's in the last 5 balls you weigh 3 of them against any other 3 from the first two groups (which are of the standard weight) and you'll narrow it to either (1) 3 balls and knowing if the odd one is lighter/heavier, or (2) the two balls which you haven't put on the scales yet.
...case (1): weigh any two of these three balls against each other and you're done
...case (2): weigh any one of these two against one of the other balls and you're done again
II.) In case the first measurement showed that the odd ball is among the first two groups (A!=B) take three from group A and two from group B and weigh them against the 5 in group C which now you know are of the standard weight.
...case (1) If they're the same the odd one is among the other three balls that you just left out - one from group A and two from group B. Comparing the two from group B will tell you the answer (you'll also have to use whether in the very first measurement A was lighter or heavier than B).
...case (2) Let's say that in the first measurement group A was lighter than group B (the following logic works the same if it were the other way).
Then if the 5 balls from A&B are lighter than the 5 good ones, the odd one is among the three from group A, and it's lighter than the rest. Otherwise it's among the two from group B and it's heavier. The last measurement will narrow that down exactly as as in case (I).
Something you neglected to tell us is that we should know if the odd ball out is heavier or lighter than the others.
X
X- The wording was deliberate I think. But tbh you dont need to know whether the odd one is lighter or heavier. Just that the rest are all identical in weight and that the odd one is of a different weight.
My puzzle is shamelessly copied from a source, but I wont disclose where until the answer comes.
At a local fairground, there are 10 two-seater cars attached to the ferris wheel. The ferris wheel turns so that one car rotates through the exit every minute. The wheel begins operating at 10 in the morning and closes 30 minutes later. Whats the maximum number of people that could have gone round on the ferris wheel in that time? :hmmm:
Nope, you do not have this information ahead of time and in some cases you would find it at the end, in some cases you won't.Quote:
Originally Posted by xman
If you knew the problem is pretty trivial. In fact in that case you can find the odd one among 27 balls with three measurements.
What is the definition of closing - when it's completely empty or when it stops accepting new riders. In the later case do partial rounds count?Quote:
Originally Posted by sidneykidney
Also does it mean that each car rotates trough the exit every minute or there is a car passing through the exit every minute - that's factor of 10 in speed you know :)
The ride closes at 10.30. At 10.30 the ride operator leaves the ride and goes home. Is that clearer?
Any more than that and you'll be asking me for the answer. Take a guess gugi. If you get it wrong you can always guess again.
42 people
Assuming there is a couple in the seat at 10, 60? or is it 58?
Well I don't like to guess when I can just compute the correct answer, but that would necessitate unambiguously defined problem :)Quote:
Originally Posted by sidneykidney
So presuming each car passes through the exit once a minute, and not that 'a car' passes through the exit once a minute, the first car would've passed 30 times during the time of operation and the other 9 cars would've passed 29 times each. Which means the maximum number of people who could've had a full ride is 2*(30+9*29)=582.Quote:
Originally Posted by gugi
Lee is correct. The thing which initially caught me out is that you have to have everybody off by 10.30. When the ride closes and the operator goes home he wont leave people sitting on the ride! Can you show your working Lee? Or did you google the answer? :p
You must be glad I'm not the operator...Quote:
Originally Posted by sidneykidney
And this means the speed of the wheel is the slow one where each car passes by the exit once every ten minutes... I guess it's the old folks roundabout.