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# Pouring Water VII.

## Question

Three Jars contain 19,13 abd 7 liters respectively. The first is empty and the others are full. None of the vessels is graduated. How can one measure out 10 liters, using no other vessels, by pouring fluid from one into another?

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or shall i post spoiler.

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so we got:

0/19

13/13

7/7

??

if thats the case i'll get right on it. i like these puzzles!

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yes unreality. your on a right track.

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can the jars hold as much as they want?

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???

19 liter jar can hold only 19 liters no more. However you have total of 20 liters. You cant get more or dump any out.

does that help?

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k thanks!

Jar A: 0/19

Jar B: 13/13

Jar C: 7/7

B into C

Jar A: 13/19

Jar B: 0/13

Jar C: 7/7

C into B

Jar A: 13/19

Jar B: 7/13

Jar C: 0/7

A into B until B is full

Jar A: 7/19

Jar B: 13/13

Jar C: 0/7

B into C

7/19

6/13

7/7

C into A

14/19

6/13

0/7

B into C

14/19

0/13

6/7

A into B

1/19

13/13

6/7

B into C

1/19

12/13

7/7

C into A

8/19

12/13

0/7

lol you've probably realized by now i'm confused as hell...

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hahaha yes you are

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rookie1ja ---- Martini .... did you guys solve it yet?

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There's probably an easier way, but the below appears to work <!-- s:D --><!-- s:D -->

First figure is the 19 jar, then the 13, then the 7

0 - 13 - 7

7 - 13 - 0

19 -1 - 0

12 - 1 - 7

12 - 8 - 0

5 - 8 - 7

5 - 13 - 2

18 - 0 - 2

18 - 2 - 0

11 - 2 - 7

11 - 9 - 0

4 - 9 - 7

4 - 13 - 3

17 - 0 - 3 Correction

17 - 3 - 0

10 -3- 7

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rookie1ja ---- Martini .... did you guys solve it yet?

just wanted to post my solution

of course, there is a shorter way ... but I figured this one out without looking at other posts

btw, I wanted to get somewhere 4 or 3 liters (then it's easy)

0 - 13 - 7

7 - 13 - 0

7 - 6 - 7

14 - 6 - 0

14 - 0 - 6

1 - 13 - 6

1 - 12 - 7

8 - 12 - 0

8 - 5 - 7

15 - 5 - 0

15 - 0 - 5

2 - 13 - 5

2 - 11 - 7

9 - 11 - 0

9 - 4 - 7

16 - 4 - 0

16 - 0 - 4

3 - 13 - 4

3 - 10 - 7

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rookie1ja, slight error in your last step, It should read 3 10 7 not 3 10 4. But you still got it in fewer steps than me well done

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Nice!

I think normdeplume got it in fewer steps(16). and rookie1ja in 19 steps.

I came up with same solution as normdeplume. I dont think theres an easier way. However if there is plz post it.

peace...

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rookie1ja, slight error in your last step, It should read 3 10 7 not 3 10 4 . But you still got it in fewer steps than me well done

typo corrected

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0 - 13 - 7

7 - 13 - 0

19 -1 - 0

12 - 1 - 7

12 - 8 - 0

5 - 8 - 7

5 - 13 - 2

18 - 0 - 2

18 - 2 - 0

11 - 2 - 7

11 - 9 - 0

4 - 9 - 7

4 - 13 - 3

17 - 0 - 0

17 - 3 - 0

10 -3- 7

look at the last four steps:

4 - 13 - 3

17 - 0 - 0

17 - 3 - 0

10 -3- 7

am I the only person who caught this?

the OP said clearly there was 20 liters total. No pouring out into a sink and no getting water from anywhere.

I think you meant to do this:

4 - 13 - 3

17 - 0 - 3

17-3-0

10-3-7

that works. good job tho!

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the OP said clearly there was 20 liters total. No pouring out into a sink and no getting water from anywhere.

That's true. The OP stated "How can one measure out 10 liters, using no other vessels, by pouring fluid from one into another". That means that dumping any out or adding any would break the condition "by pouring fluid from one into another".

I think you meant to do this:

4 - 13 - 3

17 - 0 - 3

17-3-0

10-3-7

Nice!

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unreality ... we or at least i caught it, however it was obvious that it was typing error

peace...

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Can be easier:

0/19 - 13/13 - 7/7

7/19 - 13/13 - 0/7

19/19 - 1/13 - 0/7

12/19 - 1/13 - 7/7

12/19 - 8/13 - 0/7

10/19 - 10/13 - 0/7

2 10s is better than 1

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Can be easier:

0/19 - 13/13 - 7/7

7/19 - 13/13 - 0/7

19/19 - 1/13 - 0/7

12/19 - 1/13 - 7/7

12/19 - 8/13 - 0/7

10/19 - 10/13 - 0/7

2 10s is better than 1

On the last pour, how do you know when to stop?

If you know when to stop, why not just pour 3 immediately?

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initial:

A=0/19

B=13/13

C=7/7

C into A:

A=7/19

B=13/13

C=0/7

B into A:

A=19/19

B=11/13

C=0/7

A into C:

A=12/19

B=11/13

C=7/7

C into B:

A=12/19

B=13/13

C=5/7

A ito C:

A=10/19

B=13/13

C=7/7

Jar A has 10

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If A is 19

and B is 13

and c is 7

Just pour half of jar b and half of jar c into a, that would solve the puzzle.

Then A is 10

and B is 6.5

and C is 3.5

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initial:

A=0/19

B=13/13

C=7/7

C into A:

A=7/19

B=13/13

C=0/7

B into A:

A=19/19

B=11/13

C=0/7

A into C:

A=12/19

B=11/13

C=7/7

C into B:

A=12/19

B=13/13

C=5/7

A ito C:

A=10/19

B=13/13

C=7/7

Jar A has 10

I think you picked up 10 extra liters somewhere... You start with 20 Liters, but you end up with 30.

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This is how I did it...

Start:

A 0/19

B 13/13

C 7/7

Everything into A leaves 1 behind

19/19

1/13

0/7

Fill C from A

12/19

1/13

7/7

Empty C into B

12/19

8/13

0/7

Fill C with A

5/19

8/13

7/7

Fill B with C

5/19

13/13

2/7

(2 Steps here) B into A and then C into B

18/19

2/13

0/7

Fill C from A

11/19

2/13

7/7

C into B

11/19

9/13

0/7

Fill C from A

4/19

9/13

7/7

Fill B from C

4/19

13/13

3/7

B into A

17/19

0/13

3/7

C into B

17/19

3/13

0/7

Fill C from A

10/19

3/13

7/7

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I think somebody posted this solution, but this is what I thought of when I saw the puzzle.

Pour half of the other two jars into the 19 liter jar. Half of 13 + half of 7 = 10. You can be certain that you poured exactly half from each jar when the fluid touches the bottom tip of the jar and the top back of the jar at the same instant.

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I dont think you can pour halfsies out..... the 1st half out, I can see, but how do you know you have EXCATLY half out of the second jar......dont you still have the liquid from the 1st jar??

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I dont think you can pour halfsies out..... the 1st half out, I can see, but how do you know you have EXCATLY half out of the second jar......dont you still have the liquid from the 1st jar??

Geometrically speaking, if you "cut" the jar in half diagonally with the line of liquid (as you pour, the liquid will touch the tip of the jar that you are pouring from and the top back of the jar at the same time), you have poured out half of the liquid. I suppose my solution would only stand if the jars were symmetrical, which is not specified by the problem. If they were some odd shape, you could not be certain that half of the liquid was emptied.

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