Your vehicle's 7-liter radiator runs totally out of water. Thankfully you are next to a river. Your trunk has an empty 11-liter can and an empty 5-liter can. You take the cans to the river to utilize its water and then come back to your vehicle with exactly 7 liters of water to fill your radiator. How did you do it?

(There are no markings on the cans showing fractional amounts of filling, by the way.)

Originally posted by Paul Dirac II Your vehicle's 7-liter radiator runs totally out of water. Thankfully you are next to a river. Your trunk has an empty 11-liter can and an empty 5-liter can. You take the cans to the river to utilize its water and then come back to your vehicle with exactly 7 liters of water to fill your radiator. How did you do it?

(There are no markings on the cans showing fractional amounts of filling, by the way.)

I Fill the 11 l can using the five liter can 3 times. In the 5l can remain 4l.
I empty the 11 l can
and fill it beginning with the 4 l in the small can. in that remain 3 l.
I empty the big can
and fill it again starting with the 3 l. in the small can remain 2 l.
I empty the can and fill in the 2l, and one more helping of 5 liters.

I fact I would estimate to get about 8 l in the big can and discard the rest at the road π

If I show each step as an ordered pair, the first number being how much water is in the big can and the second number being how much water is in the small can, Ponderable's sequence is:

0 0
0 5
5 0
5 5
10 0
10 5
11 4
4 0

After that a 3 comes into it somehow, but I am not seeing how. π

Fill 11 liter can
Fill 5 liter can from 11 liter can (6 liters in 11liter can)
Dump 5 liter can
Fill 5 liter from the 6 liters remaining in 11 liter can ( 1 liter remains in 11 liter can)
Dump 5 liter
fill 5 liter with 1 liter remaining in 11 liter can
fill 11 liter can
fill 5 liter can with 1 liter inside from 11 liter can
7 liters remain in 11 liter can.

However, practically you should find and fix the leak, then fill up the 11 liter can, take it to your vehicle, and fill the 7 liter radiator...

Originally posted by Paul Dirac II If I show each step as an ordered pair, the first number being how much water is in the big can and the second number being how much water is in the small can, Ponderable's sequence is:

0 0
0 5
5 0
5 5
10 0
10 5
11 4
4 0

After that a 3 comes into it somehow, but I am not seeing how. π

the sequence would be:

0 0
0 5
5 0
5 5
10 0
10 5
11 4
0 4 (emptied the big can)
4 0
4 5
9 0
9 5
11 3
0 3 (emptying the big can)
3 0
3 5
8 0
8 5
11 2
0 2 (emptying the big can third time)
2 0
2 5
7 0

Originally posted by joe shmo Are these types of problems formalized in mathematics somewhere?

There may be some branch of math that treats this sort of thing systematically. All I can tell you is this was problem 59 in a book of puzzles, with the other puzzles not being of this nature. The book provides the answers in the back. The book's answer for 59 does not try to generalize the procedure at all.

Originally posted by joe shmo Are these types of problems formalized in mathematics somewhere?

Well my idea was to to exploit the 1 Liter difference to obtain the required amount. I think as long as there is a one Liter difference you can solve for any amount (given the right size of the bigger vessel).
Joe's solution was far more elegant, since it exploited the difference more effective.

Originally posted by Ponderable Well my idea was to to exploit the 1 Liter difference to obtain the required amount. I think as long as there is a one Liter difference you can solve for any amount (given the right size of the bigger vessel).
Joe's solution was far more elegant, since it exploited the difference more effective.

Of course you could just fill the 11 liter can, fill the radiator, then have 4 liters to drinkπ

I just threw that in because I am dying of thirstπ Too much beer makes me thirsty.....