If the size 3, 5, and 6 containers are A, B, and C, then dump C into A, then B into C, then A into B, then C into A, then B into C, which will be full of 72 degree water. Thanks for your site, Earl Gose
Capacity 3 5 6
Start Volume 0 5 6
Temp 0 50 90
Step 1 Volume 3 2 6 Pour bucket 2 into bucket 1 until bucket 1 is full
Temp 50 50 90
Step 2 Volume 3 5 3 Pour bucket 3 into bucket 2 until bucket 2 is full
Temp 50 74 90
Step 3 Volume 0 5 6 Pour bucket 1 into bucket 3 until bucket 3 is full
Temp 0 74 70
Step 4 Volume 3 5 3 Pour bucket 3 into bucket 1 until bucket 1 is full
Temp 70 74 70
Step 5 Volume 3 2 6 Pour bucket 2 into bucket 3 until bucket 3 is full. Bucket 3 has 6 litres of water at 72 degrees
Temp 70 74 72
Trotter, Alasdair
Assume containers are symmetric.
Tilt the 5 lit letting water fall out until water level goes from a bottom edge to opposite top edge: that will empty exactly half; is that allowed?
Assuming it is allowed.
We have 250 lit left in the 5 lit;
fill it from 6 lit: we get 500 lit @ 70 degrees;
is that what's being asked?
I can then fill the 3 lit with this 70 degree mix;
empty half the 3 lit into the 6 lit: 150 @ 70 + 350 @ 90 = 84 degree mix; that all "within the rules" ?
Thanks, Ed.
Me: Doctor, got my leg broken in 3 places...
Doctor: stay away from those places !
Denis Borris
Well, my "half pouring" trick probably not allowed;
was able to do it in 5 moves not using it:
(A/B/C = 3/5/6 containers)
0: before 1st go : 0/00, 5/50, 6/90
1: fill A from B : 3/50, 2/50, 6/90
2: fill B from C : 3/50, 5/74, 3/90
3: fill C from A : 0/00, 5/74, 6/70
4: fill A from C : 3/70, 5/74, 3/70
5: fill C from B : 3/70, 2/74, 6/72
I still have a question on the other part of puzzle:
are we to get as many integer temperatures from 51 to 89 as possible?
are my "70" and "74" above 2 of them ?
Denis Borris
Fill the 3 liter with 50 and top off the five liter with 90 leaving:
3l full 50
5l full 74
6l 3 lit @ 90
Put all the 50 in the 6 liter w/ the 90
3l EMPTY
5l full 74
6l full 70
Put 3 lit of 70 in the 3l container and top off the six liter with 70 leaving
3l full 70
5l 2 lit @ 74
6l full 72
Rick Kline
After 20 minutes of playing, I can obtain the following temperatures: (not
including the starting temperatures!)
60 deg: B->A;B emptied;C->B;A->C
61
62 deg: C->A;B->C;C->B
63
64
65
66 deg: C->A;B->C;C->B;B->C
67
68 deg: C->A;B->C;C->B;A->C;B->A;A emptied;B->A;C->A
69
70 deg: C->A;B->C
71 deg: C->A;B->C;C->B;A->C;B->A;B emptied;A->B;C->A;B->C
72 deg: B->A;C->B;A->C;C->A;A emptied;B->A;A->C
73
74 deg: B->A;C->B
75
76 deg: C->A;B->C;C->B;A->C;B->A;A
emptied;B->A;C->A;A->C;C->A;A->B;C->B;B->A;A->C
77
78 deg: C->A;B->C;C->B;A->C;B->A;A emptied;B->A;C->A;A->C
79
80 deg: C->A;B->C;C->B;A->C
81
82 deg: B->A;A emptied;C->A;B->C;C->B;A->C;B->C
I had written the attached program for a friend - you can configure the defaults if you wish to try other arrangements. (It requires mscomctl.ocx in your system directory. [MS common controls OCX] If you don't have it, let me know and I can send you it. (1MB zipped to 450KB))
Alan O'Donnell
Ed,
I hope you managed to get my 'water' program running ok.
FYI, I have a new high score of 85 degrees, operations to achieve 73/75, and
a shorter 82.
I don't think anything higher than 85 is possible.
Alan O'Donnell.
60 deg: B->A;B emptied;C->B;A->C
61
62 deg: C->A;B->C;C->B
63
64
65
66 deg: C->A;B->C;C->B;B->C
67
68 deg: C->A;B->C;C->B;A->C;B->A;A emptied;B->A;C->A
69
70 deg: C->A;B->C
71 deg: C->A;B->C;C->B;A->C;B->A;B emptied;A->B;C->A;B->C
72 deg: B->A;C->B;A->C;C->A;A emptied;B->A;A->C
73 deg:*B->A;C->B;A->C;B->A;B emptied;C->B;A->C
74 deg: B->A;C->B
75 deg:*C->A;B->C;C->B;B->C;B
emptied;A->B;C->A;B->C;A->B;C->A;B->C;A->B;C->A;B->C
76 deg: C->A;B->C;C->B;A->C;B->A;A
emptied;B->A;C->A;A->C;C->A;A->B;C->B;B->A;A->C
77
78 deg: C->A;B->C;C->B;A->C;B->A;A emptied;B->A;C->A;A->C
79
80 deg: C->A;B->C;C->B;A->C
81
82 deg:*B->A;A emptied;C->A;C->B;A->C;B->C
83
84
85 deg:*B->A;B emptied;C->B;A->C;C->A;B->
I've given up on this one now.
On top of my previous submission, the following includes:
1) Continuous 'solution' for 65 deg.
2) A 7-step 71 deg.
3) A 3-step 82 deg.
4) A 5-step 85 deg.
Note all of these were done by hand using the 'solver' program I wrote and
sent to you the other day.
All the best
Alan O'Donnell.
60 deg: B->A;B emptied;C->B;A->C
61
62 deg: C->A;B->C;C->B
63
64
65 deg:*C->A;A emptied;B->C;C->B;[B->C;C->B repeated to oo] tends to exactly
65 deg.
66 deg: C->A;B->C;C->B;B->C
67
68 deg: C->A;B->C;C->B;A->C;B->A;A emptied;B->A;C->A
69
70 deg: C->A;B->C
71 deg:*C->A;B->C;C->B;A->C;B->A;C->B;A->C
72 deg: B->A;C->B;A->C;C->A;A emptied;B->A;A->C
73 deg:*B->A;C->B;A->C;B->A;B emptied;C->B;A->C
74 deg: B->A;C->B
75 deg:*C->A;B->C;C->B;B->C;B
emptied;A->B;C->A;B->C;A->B;C->A;B->C;A->B;C->A;B->C
76 deg: C->A;B->C;C->B;A->C;B->A;A
emptied;B->A;C->A;A->C;C->A;A->B;C->B;B->A;A->C
77
78 deg: C->A;B->C;C->B;A->C;B->A;A emptied;B->A;C->A;A->C
79
80 deg: C->A;B->C;C->B;A->C
81
82 deg:*B->A;C->B;B->C
83
84
85 deg:*C->A;B->C;B emptied;C->B;A->C
Dear Sir/Madam, Iam here by sending you the answer for the puzzle which was sent by Cihan Altay ( http://www.mathpuzzle.com/ ). Given below is the answer for the same. 1. we have to first take 5 litter containor and fill in the 3 liter containor. Now (I) The 3 liter containor will be 50 Degree of 3 liters. In 5 litre container, 50 degree of 2 litres will be remaining. 2. now we have to fill from 6 litres containing to 5 litre container. (II) now in 5 litre container water content is 74 degrees of 5 litres and in 6 litres contaniner, 90 degrees of 3 litres will be remaining. 3. we have to fill the 6 litres container from 3 litre container. (III) now in 6 litres container the water will be full with 70 degrees. [(90 + 50) / 2 ] degree. 4. Now fill the empty 3 litre container from 6 litres container. 5. fill the 6 litres container from 5 litres. now we have remaining 2 litres in 5 litre container. (IV) Now in 6 litres container water will be full of 72 degree[(70 + 74)/2 ]=72 degree. *** = 72 degrees in 6 liters of 6 liters continer . ( which is the answer )
In the three vessel problem, that
you posted on your site, I forgot
to add an additional step!
The right procedure should be:
1.- Pour 3 lt from vessel B into vessel A,
2.- Pour 3 lt from vessel C into vessel B,
3.- Pour 3 lt from vessel A into vessel C,
4.- Pour 3 lt from vessel B into vessel A
and empty vessel B,
5.- Pour 3 lt from vessel A back into
vessel B,
6.- Pour 3 lt from vessel C into vessel A,
7.- And finally pour 3 lt from vessel B
into vessel C to obtain 6 lt of liquid
of the desired temperature.
Thanks for the problem, it was fun!
Luis Rodriguez
I have two solutions to the puzzle water mixing puzzle sent to you by Cihan Altay:
Solution 1:
The heat in the 5 liter container = 5*50 = 250 units; that in the 6 liter = 6*90 = 540 units. So there are a total of 790 units of heat shared between 11 liters or 790/11 units of heat per liter. This equates to nearly 72 units of heat per liter (exactly 71.8181 units per liter), or very nearly 72 degrees.
If this small error is permitted then I would pour all of the water from the 6 liter container into the 5 liter container so that the mixture overflows. Provided the water mixes thoroughly during the pouring process the temperature of the liquid that ends up in the 5 liter container will be (very nearly) 72 degrees.
Solution 2:
Do nothing. If the temperature of the surrounding is more than 72 degrees then the 5 liters will warm up to that temperature; if the surroundings are cooler then the 6 liters will cool to that temperature.
Note that any temperature between 50 and 90 degrees can be obtained. If the surroundings are cooler than 50 degrees or warmer than 90 then more integral solutions are possible than you ask for in your Mega-Puzzle.
The neatness of both of these solutions is that the 3 litre container is not required.
Roy Lambert
You can get to 72 degrees by the following moves:
B=>A C=>B A=>C B=>A C=>B A=>C
where A = 3lt cup, B = 5lt & C = 6lt.
At the end C contains 2.7lt of 50 degree water + 3.3lt of 90 degree water. There are other solutions but this is the shortest I found.
The sequence of temperatures & amounts in each cup are:
A temp Empty 50 50 Empty 74 74 Empty
B temp 50 50 74 74 74 71.6 71.6
C temp 90 90 90 70 70 70 72
A liters 0 3 3 0 3 3 0
B liters 5 2 5 5 2 5 5
C liters 6 6 3 6 6 3 6
Moves=> B=>A C=>B A=>C B=>A C=>B A=>C
Regards, Lorne Anderson
puzzle is
6Lt container is full 90 deg water, 5 Lt 50 deg water and last one 3 Lt container empty.
Solution will be
if we named 6 Lt container A ,5 Lt container B and empty one C
1-) container B pour the water and fill container C up and container A pour and fill it up water to container B
now we have Container A 3 Lt 90 deg water
B 5 Lt 74 deg water
C 3 Lt 50deg water
2-) Now pour the water of container C to the Container A
we have container A 6 Lt 70 degree water
3-) pour the container A to the container C
so we have container B 5Lt 74 deg water and Container A 3 Lt 70 degree water
4-) fill container A up with container B so we have 6 Lt 72 degree water in Container A,Container B is 2 Lt 74 deg water and container C is 3 Lt 70 degree water.
I hope understandable that explanation
Murat Dedeoglu
Dear Sir/Madam,
Iam here by sending you the answer for the puzzle which was sent by Cihan Altay
( http://www.mathpuzzle.com/ ). Given below is the answer for the same.
1. we have to first take 5 litter containor and fill in the 3 liter containor.
Now
(I) The 3 liter containor will be 50 Degree of 3 liters. In 5 litre
container, 50 degree of 2 litres will be remaining.
2. now we have to fill from 6 litres containing to 5 litre container.
(II) now in 5 litre container water content is 74 degrees of 5 litres and
in 6 litres contaniner, 90 degrees of 3 litres will be remaining.
3. we have to fill the 6 litres container from 3 litre container.
(III) now in 6 litres container the water will be full with 70 degrees.
[(90 + 50) / 2 ] degree.
4. Now fill the empty 3 litre container from 6 litres container.
5. fill the 6 litres container from 5 litres. now we have remaining 2 litres
in 5 litre container.
(IV) Now in 6 litres container water will be full of 72 degree[(70 + 74)/2
]=72 degree.
*** = 72 degrees in 6 liters of 6 liters continer . ( which is the answer )
shanmugam av