wolfgang
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Posts posted by wolfgang
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I`ll go with the 5th. option.
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Nice solve, wolfgang!
Thank you dear Bonanova
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Can I ask some questions?
1- Can any of the prisoners catch the mice?
2- Is the speed of the mice fix?
3- Is it possible to hang any paper around the mice`s neck?
4- Do each one of the prisoners have a watch?
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If we have an emty cylindrical glass on a flat table with a dice inside it , naturally, it will contain an amount of air inside it.
Now , if we invert this glass upside down on the table, leaving the dice still inside it but landing on the table , will the amount of air inside the glass changes?
If yes, would it be more or less, and why?
If no...why?
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1
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1- move second hand CCW 48 marks.
2- move second hand CW 41 marks.
and it will be at 2 O`clock -
I do not know it is Mr. Stone who was in prison..but you gave up the answer..so you must be in prison.
Good to see you again wg.
thank you dear...I give up....the trap didn`t work
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Mr. Bone and Mr. Stone are friends....one of them is now in prison.
What do you think....which person is now in prison?
Note: any one knowing the right answer will be put in prison instead of Mr. Stone.
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Hi friends...I missed u all... hope you are doing well....
As usual...I`ll start with a very easy puzzle...enjoy it....thanks
If
6+27=195
8+12=236
9+18=270
11+13=321
find... 10+10=?
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normal distribution curve, one should refer to swap if his (x) value was within the range of least 3%. -
Great to have you back!
nomatter.JPGThanks dear T.S.L.F. , sorry....but you changed their arrangement....which should be( A,B,C,D) and not ( B,A,C,D).
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Hi dear friends...I missed you so much....I love you all...
I have today a nice puzzle for you...hope to enjoy solving it
..We have 4 beakers( A,B,C, and D)
(A) can take only 3 gallons of water,(B) can take 4 gallons...they are both empty at start position.
© and (D) have the same size , each can take exactly 5 gallons, both are full with water to start with.
Use these beakers only to make an ascending or descending order of water amount in each beaker...
i.e. , either...1 gallon, 2,3,and 4 gallons,
or 4 gallons, 3,2 and one gallon...
Have a nice time..
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If we took 2 balls in each side, and we are told they are equal, so we should expect to have one out of 16 possibilities:
A+B= C+D
A+B= D+C
B+A= C+D
B+A= D+C
and so on...
for second weigh..
A against B....we should expect to have one of 8 results as( >) , or one of 8 results as( < ), and the 3rd weigh should be..
C against D...
now we can determine the two heavy balls and the two light ones
with 4th weigh, between the two heavies... and 5th between the two others, we can arrange them. -
Even if she was born on Feb.29.
she should have at least $1800....
but to find only $500!!
is really strange
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but my first pass through to find both fakes, assuming that the balance scale simply tells you "heavier" or "lighter" and that you can't measure HOW MUCH heavier or lighter is 5 weighings...I have a couple different approaches using 5 weighings, but will look for fewer.
Presume it tells you "equals" as well.
yes...equals..is also possible
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You have 8 coins that all look exactly the same, two of them are fake,one of these two fake coins is 0.1gm more than a normal coin, whereas the second coin is 0.1 gm less than a normal coin.
With howmany weighings can you find them out using a balance scale?
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1
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B will win with 51 to 49
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Didn`t find one yet
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S...because it is unique
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Why there are only 7 letters in the 1st row?
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1.N d6 ,with the next move to f7 then to h8
.....if black,N e4 to block f7
2.N c6 with the next move to b8...the black can not take it
...if he tried to cover the b8 with N a3...then
3. Nx e6!-
1
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when you cut the two circles into half, you can put these 4 pieces on a sheet of paper and make a square...
Make two circle with equal diameter. Cut both of them into half, so that four halves are available. Now.two halves may be put together in such a way that the curved surfaces touche each other and diameter sides are parallel to each other - say you get a fig. 'A'. Now put two halves of other circle over fig. 'A' in such a way that diameter sides are at 90 degree to the dimeter sides of fig. 'A' and the curved surfaces touch each other at the same point where those of first circle are in touch. You will get a Square with a flower design inside.
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both of them are correct...and still there is another method
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I take it in such a way that 4 ships are connected to a 5th ship from 4 sides to give it the energy all the trip ,i.e. only the central ship will be functioning and the others adhering to it are as energy suppliers only.
Yes...thats right...Thanks
If the fuel is used for guidance and anti-explosibility measures only, and not for propulsion; then all five spaceships could be flung from the moon towards Mars starting at the same time. All the ships would be tethered in such a way that adjusting the course of one would alter the course of the fleet as a whole. For the first quarter of the trip, Spaceship #1 does all of the course correction. For the second quarter, Spaceship #2; and so on until landing on Mars. Once on Mars, Spaceship # 5 redistributes it's fuel equally between all of the spaceships to power their lasers and defeat the Martians.
Dear wolfgang, the solution that you consider correct suffers from two problems: 1) It violates common understanding of thermodynamics and 2) even if we allow such violation, it does not satisfy the conditions in your original post. Here is why
From your original post, each ship needs to make constant trajectory corrections since "without fuel the spaceship will miss its direction". The solution that you accept states that 4 space-ships can "piggy-back" on the course correction ability of 1 ships. That is, the ships would be tethered in such a way that adjusting the course of 1 ship would alter the course of the whole.
Of course, a simple understanding of physics would indicate that we don't really save any fuel if we allow 1 ship to change the course for all 5. If a single ships needs to make a minor course correction, it would need to provide enough thrust (energy) to move its mass however much required. If a single ship is tethered to 4 other ships, Newton's equation (F=ma) implies that the navigating ship would need 5 times the energy to alter the current course. So, in reality, tethering 5 ships together would make the single navigating ship burn fuel 5 times as fast as it changes course.
That said, even if we allow such violation, how do you suppose to get the ships home. From your OP, "each spaceship has a fuel capacity to allow it to fly exactly 1/4 way to Mars" and "Each spaceship must have enough fuel to return safe to the base space station". Even allowing for the thermodynamics violation, all your 5 ships would end up on Mars with 1 tank left, which is only good enough for 1/4 of the way back to the moon base.
While I understand that lateral and out-of-box thinking are valued around here, in general such solutions have to be internally consistent (and avoid violating explicit conditions set forth in the original post). Also, in this forum, we obey the law of thermodynamics.
Three cars
in New Logic/Math Puzzles
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