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| Do you enjoy riddles? |
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| Total Votes : 5 |
Lintra
Elf Friend
Joined: 23 Apr 2002
Posts: 9448
Location: Bermuda, the triangle place with SANDY BEACHES |
quote:
Originally posted by Beebelo
Flip the first switch on, wait
Flip the first switch off
Flip the second switch on
Enter the cave
If the light is on, then the second switch is the right one
If the light is off, but the bulb is hot, choose the first switch
If the light is off, but the bulb is cold, choose the third switch
PERFECT. I really wish I had thought of that  |
Thu May 16, 2002 7:08 pm |
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Yog Sothoth
The All in One
Joined: 01 May 2002
Posts: 1086
Location: Between Space and Time |
quote:
Originally posted by Beebelo
Flip the first switch on, wait
Flip the first switch off
Flip the second switch on
Enter the cave
If the light is on, then the second switch is the right one
If the light is off, but the bulb is hot, choose the first switch
If the light is off, but the bulb is cold, choose the third switch
SURE! Pure genius!  Good Work!
_________________
=Member of the NonFlamers' Guild=
=Memberof the 6thHouse=
=Member of the SportsFans Club=
=Ex-Baron of the RPGDot Shadows= |
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Thu May 16, 2002 10:44 pm |
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Michael C
Black Dragon
Joined: 09 Jul 2001
Posts: 1595
Location: Aarhus, Denmark |
Yep good work on the lightswitch problem.
The "timing is everything" riddle (Train problem). I calculated that he had to be walking 50 Min., but I'm not sure if it's right, or how to prove it. Any solutions??? Math problem for sure! |
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Fri May 17, 2002 7:25 am |
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Beebelo
Village Dweller
Joined: 14 May 2002
Posts: 5
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For Ties that Bind (string problem) how's this for an answer...
Light both ends of 1 string (=30 mins)
When that one is done, fold the 2nd string in 1/2 and light all 3 ends (=15 mins) [technically 4 ends, as the middle will separate]
Still working on the timing one...
_________________
"It's not denial, I'm just very picky about which reality I choose to live in" Calvin&Hobbes |
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Fri May 17, 2002 3:11 pm |
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Lintra
Elf Friend
Joined: 23 Apr 2002
Posts: 9448
Location: Bermuda, the triangle place with SANDY BEACHES |
quote:
Originally posted by Beebelo
For Ties that Bind (string problem) how's this for an answer...
Light both ends of 1 string (=30 mins)
When that one is done, fold the 2nd string in 1/2 and light all 3 ends (=15 mins) [technically 4 ends, as the middle will separate]
Still working on the timing one...
The 1/2 hr works, but the 15minutes doesn't. What if the burining time is distributed as follows:
first quarter: 1 min
second quarter: 1 min
third quater: 1 min
Fourth quarter 57 minutes.
When the fourth quarter finished buringing you would have timed 28.5 minutes! But I like the folding idea, you are on to something there. |
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Fri May 17, 2002 4:20 pm |
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Michael C
Black Dragon
Joined: 09 Jul 2001
Posts: 1595
Location: Aarhus, Denmark |
Beebelo: If you light the first string from both ends, and at the same time light the second string from only one end. When the first string is finished (30Min), you light the other end on the second string too, and the second string will be burned out 15 min. after that, giving a total of 45 min. |
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Tue May 21, 2002 8:17 am |
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Michael C
Black Dragon
Joined: 09 Jul 2001
Posts: 1595
Location: Aarhus, Denmark |
The "timing is everything" riddle (Train problem).
Okay here is how I reason for the solution on 50 Min.
Normally they meet at the train station at 5.00 PM, but at the next day he arrives at 4.00 o,clock and starts walking towards home, and get picked up on the road and arrives at home 20 Min. before as usual. The car that start from home must then drive 10 Min. shorter each way this day (2 x 10Min. = 20 min.), meaning the car meets the walking person at 5.00 - 10 min. = 4.50 o,clock. The person starts walking at 4.00 o,clock, so he had to be walking 4.50 - 4.00 = 50 Min.
Elementary! |
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Tue May 21, 2002 9:37 am |
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Yog Sothoth
The All in One
Joined: 01 May 2002
Posts: 1086
Location: Between Space and Time |
quote:
Originally posted by Michael C
Beebelo: If you light the first string from both ends, and at the same time light the second string from only one end. When the first string is finished (30Min), you light the other end on the second string too, and the second string will be burned out 15 min. after that, giving a total of 45 min.
 WOW i think it's the perfect solution! Good work!
_________________
=Member of the NonFlamers' Guild=
=Memberof the 6thHouse=
=Member of the SportsFans Club=
=Ex-Baron of the RPGDot Shadows= |
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Tue May 21, 2002 9:49 am |
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Lintra
Elf Friend
Joined: 23 Apr 2002
Posts: 9448
Location: Bermuda, the triangle place with SANDY BEACHES |
@ Michael C - Good sol'ns!! I'd buy you a beer  if I were local
On my unofficial count, that leaves the following riddles left 2B solved:
Flight Plight (an infinite series - I can post sol'n if you'd like)
Hats off
Lazy Susan
Heads Up |
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Tue May 21, 2002 12:06 pm |
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Michael C
Black Dragon
Joined: 09 Jul 2001
Posts: 1595
Location: Aarhus, Denmark |
Plight Flight: I can get it down to 6 planes including the one that goes around the world.
3 planes take off from the start. 1/8 around the world 1 plane gives the two others 2/8 of it's fuel filling them both up again, and head back with it's last 2/8 tank of fuel.
The 2 planes continue until 2/8 (1/4) around the world where 1 plane gives 2/8 of his fuel to the other, giving the other a full tank, and barely self return with it's remaining 4/8 tank.
The only plane left with a full tank at 2/8 around the world can now go up to 6/8 around the world before it needs refuelling. So at 4/8 around the world 3 other planes start from the base, heading the other way around the world to meet the World Touring Airplane (WTA for short).
When WTA reach 5/8, the 3 other planes have flown 1/8 around the world, and one of them give each of the two others 2/8 fuel, and returns to base with it's remaining 2/8 fuel.
When WTA reach 6/8 around the world, it meets the other two planes, which each share 2/8 of fuel to WTA, so all 3 planes have 1/2 full planes and can all make it home to the base safely.
So 6 planes to make a world round trip for one plane! |
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Tue May 21, 2002 1:40 pm |
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Michael C
Black Dragon
Joined: 09 Jul 2001
Posts: 1595
Location: Aarhus, Denmark |
Heads up riddle: I got 20 coins with head up, and infinite number of coins with tail up. I got to make 2 piles of coins with an equal amount of head up coins, blindfolded, and can flip as many coins as desired.
If I flip all coins, I will get 20 coins with tails up and infinite number of coins with heads up. I will then make 2 equal piles of all the coins, leaving a infinite number of "head up" coins in each pile!
Last edited by Michael C on Wed May 22, 2002 1:35 pm; edited 1 time in total |
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Tue May 21, 2002 1:52 pm |
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Lintra
Elf Friend
Joined: 23 Apr 2002
Posts: 9448
Location: Bermuda, the triangle place with SANDY BEACHES |
Good sol'n on Flight Plight - I had overlooked the possibility of 3 planes to start  |
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Tue May 21, 2002 2:04 pm |
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Michael C
Black Dragon
Joined: 09 Jul 2001
Posts: 1595
Location: Aarhus, Denmark |
The lazy susan riddle:
4 positions the glasses are placed on!
P1 P2
P3 P4
4 glasses which either are (D)own or (u)p.
1. I will take glass at P1 and P4 and turn them Up
U P2?
P3? U
2. After the spin, I will take glass at P1 and P2 and turn both up. And if I have not won now, I know the following
U U
U D
However I don't know the position of the one glass (down).
3. In my third attemp, I will touch glasses at P1 and P4. If one is down, I will turn it and have won, but if both are up I will turn one of them down, giving me this picture.
U U
D D
4. In my fourth attemp I will touch glasses at position P1 and P2, if they are alike I just turn them both and win. If not I switch them both for this result:
U D
D U
5. Finally I will in my next attemp turn P1 and P4, and surely have all glasses either down or up, leaving me with 5000$. |
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Tue May 21, 2002 2:37 pm |
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Lintra
Elf Friend
Joined: 23 Apr 2002
Posts: 9448
Location: Bermuda, the triangle place with SANDY BEACHES |
quote:
Originally posted by Michael C
Heads up riddle: I got 20 coins with head up, and infinite number of coins with tail up. I got to make 2 piles of coins with an equal amount of head up coins, blindfolded, and can flip as many coins as desired.
If I flip all coins, I will get 20 coins with tails up and infinite number of coins with heads up. I will then make 2 piles of all the coins, leaving a infinite number of "head up" coins in each pile!
I am not sure this works (though it seems to). To make two piles, one of them has to be finite, unless you specify something like every other coin goes into pile one, the rest go into pile 2 ... Yes that works! Good sol'n  |
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Tue May 21, 2002 3:00 pm |
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Yog Sothoth
The All in One
Joined: 01 May 2002
Posts: 1086
Location: Between Space and Time |
From the second link in the riddle page: choose the hard riddle:
number 68. 8 BBs look alike but one is slightly heavier than the others. How can you identify the heavy BB in only two weighings on a balance scale?
(think about it)
Solution:
put 3 BB and 3 BB = if the weight is equal = i weight the aothers 2 DONE!
If the weight is unequal = take the 3 heavier and weight 1 vs 1 DONE!
Simple... 
_________________
=Member of the NonFlamers' Guild=
=Memberof the 6thHouse=
=Member of the SportsFans Club=
=Ex-Baron of the RPGDot Shadows= |
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Tue May 21, 2002 3:30 pm |
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