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This question has been bugging me for a day and finally I gave up and decided to ask the community for it so here's how it goes:

Suppose we have 3 bottles with capacities of $16,6$ and $3$ liters, all of which are full of water currently.

In what way can we move the water from a bottle to another one (spilling is also an option) that only $11$ liters of water remain in the $16$-liter bottle?

EDIT:Thanks for your comments and now I believe you may have misinterpreted my question and therefore I think I'll unmark the answer for now(which I realized was not true after a while).

EDIT2:I forgot to mention the following conditions but they pretty much could be interpreted from the question iteself:

  1. We cannot measure a bottle half-way.
  2. The only scales we have are only these 3 bottles (This one is more like an emphasize on the first condition but anyway I needed to add it)
  3. There's no reservoir where we can keep the spilled water.
  4. There are no other water resources from which we can add water.

and finally I need to point out the fact that the question may or may NOT have an answer so it's more like a "prove or disprove" kind of question.

EDIT3:Multiple transfers are allowed (Thanks to @David for pointing it out)

Thanks for your help in advance.

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    $\begingroup$ Are you allowed to refill the bottles? In that case it would be easy but I'm guessing that's not allowed? $\endgroup$
    – DRF
    Apr 24, 2015 at 8:24
  • $\begingroup$ @DRF well of course not!The only water supply you have are these 3 filled bottles. $\endgroup$
    – Arian Tashakkor
    Apr 24, 2015 at 8:27
  • $\begingroup$ Also just trivially, see this m.wolframalpha.com/input/… $\endgroup$
    – Aditya Agarwal
    Apr 24, 2015 at 8:37
  • $\begingroup$ @ArianTashakkor is the radius for each 3 bottle the same? $\endgroup$
    – ministic2001
    Apr 24, 2015 at 9:54
  • $\begingroup$ This question is unclear. It doesn't clearly state what is allowed and what isn't. Someone could just use a scale and weigh the water but it doesn't say in the question scales are not allowed. $\endgroup$
    – David
    Apr 24, 2015 at 14:56

14 Answers 14

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It is trivially possible. Just spill water from the $16$ litre bottle until it is empty. By the intermediate value theorem, there is an instant when the there is exactly $11$ litre left.

Nothing in the question indicates that this is about a construction, so an abstract existence proof will do.

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  • $\begingroup$ Intermediate value theorem doesn't exclude the non-integral real numbers. $\endgroup$
    – Aditya Agarwal
    Apr 24, 2015 at 13:43
  • $\begingroup$ @AdityaAgarwal: Sure it doesn't. You'll get $\pi^2$ litres at some point too. But that was not the question. $\endgroup$
    – Marc van Leeuwen
    Apr 24, 2015 at 13:54
  • $\begingroup$ But you cannot spill the water halfways or by a quarter, it has to be completely spilled. $\endgroup$
    – Aditya Agarwal
    Apr 24, 2015 at 13:55
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    $\begingroup$ @AdityaAgarwal On the contrary, the intermediate value theorem says that you cannot spill everything without spilling half the water first, and a quarter of the water before that, etc. (not that I want to invoke Zeno's paradox though...). By contrapositive, if one cannot spill half, one cannot spill the whole either. $\endgroup$
    – Marc van Leeuwen
    Apr 24, 2015 at 14:14
  • $\begingroup$ Yeah,in this context, it is alright! $\endgroup$
    – Aditya Agarwal
    Apr 24, 2015 at 14:15
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With a bit of lateral thinking, this is possible like this:

  1. Spill away water from the 16 liter bottle until it is half full. You can check for half-fullness by marking the water level on the side of the bottle and turning it upside down. If you spill away too much, you can refill from the 6 liter bottle.
  2. Now 8 liters remain in the 16 liter bottle, so just add the 3 liters.
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    $\begingroup$ I like this answer, but a possible issue here is that if you are very clumsy, you might keep "missing the mark" on your spilling and refilling, since you cannot test until after you have already done the addition/subtraction. (Your test, however, is very clever, and that is why I like the answer.) $\endgroup$
    – Eric Stucky
    Apr 24, 2015 at 11:02
  • $\begingroup$ In water spilling puzzles, you are not allowed to spill half the amount or any fractions. $\endgroup$
    – Aditya Agarwal
    Apr 24, 2015 at 13:42
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    $\begingroup$ This solution wont work. How would you know where the halfway mark is? Bottles usually have necks and some are even tapered so there would be no easy way to mark the middle to know when it is half full. For example, the mark might have to be at 45% of the height of the bottle but how would you know that? $\endgroup$
    – David
    Apr 24, 2015 at 14:37
  • $\begingroup$ @David - Simple. Assuming the midrange of the bottle is cylindrical, you just spill enough to ensure that the top will be in the cylindrical region both right-way-up and upside-down. Mark the height in each direction, then find the point midway between these two heights. This will be 50%. $\endgroup$
    – Glen O
    Apr 24, 2015 at 17:31
  • $\begingroup$ Again u r making assumptions. How can you tell just by looking at it that it is a "perfect" cylinder in the midrange subregion? I am imagining a particular soda bottle that is not, it is tapered pretty much all the way, possibly to look good and/or to match the contour of a persons hand better. Also there will likely be errors in marking (parallax for example). The OP didn't specify what margin of error is allowed cuz someone could argue it will "never remain" exactly $11$L and they would be right. $\endgroup$
    – David
    Apr 24, 2015 at 17:35
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If you have an extra reservoir that can contain at least 16 liters, it is possible:

  1. Start with 16-6-3-0
  2. Empty bottles of 6 and 3 into reservoir: 16-0-0-9
  3. Fill bottle of 6 from bottle of 16: 10-6-0-9
  4. Empty bottle of 6 into reservoir: 10-0-0-15
  5. Fill bottle of 6 with bottle of 16: 4-6-0-15
  6. Empty bottle of 6 into reservoir: 4-0-0-21
  7. Move content of bottle of 16 to bottle of 6: 0-4-0-21
  8. Refill bottle of 16 from reservoir: 16-4-0-5
  9. Fill bottle of 3 from bottle of 16: 13-4-3-5
  10. Fill bottle of 6 from bottle of 16: 11-6-3-5

I think an extra reservoir of 13 liters would also be sufficient, but then an extra step is needed. I do not claim that my solution is minimal.

Without an extra reservoir it is not possible, but other answers have already shown this.

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It is impossible. $16x+3y+6z=11$. Where $x$ and $y$ and $z$ are integers. By little inspection you'll observe that there are no such multiples of $6$ and $3$ and $16$ which add up to $11$. (Because adding more water is not allowed) If it is, then $(-10,3,27)$ is a possible solution. You have to have unlimited access to water to solve this.

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    $\begingroup$ +1... also, the left hand side is not divisible by $3$ while the right hand side is :-) $\endgroup$
    – Ant
    Apr 24, 2015 at 9:20
  • $\begingroup$ Yeah, thats what I did in the last step. $\endgroup$
    – Aditya Agarwal
    Apr 24, 2015 at 9:30
  • $\begingroup$ Then you only tell what will be the configuration? $\endgroup$
    – Aditya Agarwal
    Apr 24, 2015 at 9:56
  • $\begingroup$ Sorry, let me update. $\endgroup$
    – Aditya Agarwal
    Apr 24, 2015 at 10:00
  • $\begingroup$ Inspection by simple trial and error. I dont think the asker wanted the whole proof.?? And its also not worth it. $\endgroup$
    – Aditya Agarwal
    Apr 24, 2015 at 13:45
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If you ever empty the $16 L$ bottle, then you don't have enough water left in the other bottles to get $11 L$.
If you don't empty that bottle, then you can only change any amount of any bottle by a multiple of $3 L$, and so you can't get $11 L$ in this way.

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I'm going to assume that the allowable operations are:

  1. empty a bottle completely; and
  2. empty one bottle into another until the former is empty or the latter is full, whichever happens first.

Then this graph shows the shortest path to reach any reachable state (click here for a full-size version with readable labels):

graph of all states reachable from (16, 6, 3)

No state with a bottle containing 11 (or 2, 5, 8, 12, 14, or 15) liters is reachable.

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A state like $(x, y, z)$ where $x = 16+3k$ for some $k \in \mathbb{Z}$, $y \in \{0, 3, 6\}$, $z \in \{0, 3\}$ and $x+y+z \geq 11$ can only be transformed into a state satisfying the same criteria.

If the problem had a solution, we should have at some point $16+3k = 11$ for some $k \in \mathbb{Z}$, which is impossible.

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  • $\begingroup$ What are $x$, $y$, and $z$ in"a state like $(x,y,z)$"? $\endgroup$
    – JiK
    Apr 24, 2015 at 13:13
  • $\begingroup$ @JiK : they represent the quantities of water in the big, medium and small bottle respectively $\endgroup$
    – John Smith Optional
    Apr 24, 2015 at 14:08
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start with: 16 - 0 - 0
fill 16 liter bottle
pour into the 6 liter bottle
clear 6 liter bottle (10 liter left)
pour into the 6 liter bottle
clear 6 liter bottle (4 liter left)
pour into 3 liter bottle
clear 3 liter bottle (1 liter left)
pour the remaining liter into the 3 liter bottle
fill up the 16 liter bottle
pour into the 6 liter bottle
now the configuration is: 10 - 6 - 1
pour 1 liter onto the 10 liter, clear 6 liter bottle
result: 11 - 0 - 0

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  • $\begingroup$ Oh. So thats why the accepted answer said "impossible". :) $\endgroup$
    – user234126
    Apr 24, 2015 at 14:37
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You have 3 bottles. A FULL 16L bottle, an empty 6L bottle, an empty 3L bottle.

I will present the state as a comma-separated list.

I also assume all bottles are the same width.

state,16L bottle,6L bottle,3L bottle
initial,16,0,0
transfer 3L from 16L bottle to 3L bottle,13,0,3
transfer 6L from 16L bottle to 6L bottle,7,6,3
spill from 6L bottle until height same as 3L bottle (lose 3L),7,3,3
top up 6L from 16L bottle (3L),4,6,3
spill from 6L bottle until height same as 16L bottle (lose 2L),4,4,3
pour from 3L bottle back into 16L bottle,7,4,0
pour from 6L bottle back into 16L bottle,11,0,0

Unfortunately this does assume bottles have same diameter. Big assumption to make. There must be a better way.

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You can use 16 L pouring into 3L and 6L*2 If so, you can get 1L. Then use 16-6=10 And 1L get it

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This is a very naive solution, but I believe it will work with these conditions:

Each bottle is filled initially and there is no refilling

Original: 16-6-3

1) Empty the 3-liter completely. 16-6-0

2) Empty the 16- liter half way. 8-6-0

3) Fill 3-liter with 16 liter. 5-6-3

4) Fill 16-liter with 6 liter. 11-0-3

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  • $\begingroup$ Step 5, drink the 3L. $\endgroup$
    – David
    Apr 24, 2015 at 17:41
  • $\begingroup$ Is emptying a 16 liter halfway not part of the conditions? $\endgroup$
    – Gene Yllanes
    Apr 29, 2015 at 21:20
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Wow... the answers here are totally overthought and not in the spirit of the puzzle.

Fill 6, pour to 16  (6,0,0)
Fill 6, pour to 16  (12,0,0)
Fill 6, pour to 16 until full (16,2,0)
Fill 3, pour to 6 (16, 5,0)
Empty 16, pour 6 to 16 (5,0,0)
Fill 6 pour to 16 (11,0,0)
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  • $\begingroup$ OP forgot to mention that refilling is not allowed. See comments. $\endgroup$
    – Sebastian Negraszus
    Apr 24, 2015 at 20:19
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This question is still not clear. It states move water from a bottle to another one. It doesn't state moving it from/to multiple bottles like many of these answers appear to assume is ok. So with that constraint, my answer is pour approximately 3L from the 16L bottle into the 3L bottle and then just wait for the other approximate 2L to evaporate, then you will have 11L in the 16L bottle. This seems like a valid answer but many of the other ones do not because they use multiple transfers which (according tot the current wording of the question), is not allowed. The question did not state how long the 11L needs to be in the 16L bottle so even if it is for a short time, it meets the constraints (such as remaining for an instant or remaining for 1 minute...).

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enter image description here

  1. Discard all the water from $3$ liter bottle
  2. Fill it full with water from $16$ liter bottle
  3. Now, $16$ liter bottle has $13$ liter water, mark the level of the water
  4. Discard all the water from $3$ liter bottle
  5. Fill it full with water from $16$ liter bottle
  6. Now, $16$ liter bottle has $10$ liter water, mark the level of the water
  7. Divide the portion of the bottle between $10$ and $13$ mark into $3$ equal parts
  8. Fill one part with water from any one of the other two bottles
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    $\begingroup$ This wont work. The bottle may not be cylindrical and nowhere in the question did it state or imply it was. $\endgroup$
    – David
    Apr 24, 2015 at 14:45
  • $\begingroup$ @David, since it is a puzzle we need to assume some things otherwise we cannot solve it. I can create $ n$ number of puzzles without any solution! $\endgroup$
    – Vikram
    Apr 24, 2015 at 15:22
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    $\begingroup$ ok then I assume we can use a scale and I will just weigh the water so can then get exactly $11$ liters in the $16$ liter bottle. Once I know how much 3L of water weights, I know how much to spill out to remove $5$L. $\endgroup$
    – David
    Apr 24, 2015 at 15:26
  • $\begingroup$ @David, do you mean, if 3L weighs x then 5L weighs (5x)/3? yes that's possible $\endgroup$
    – Vikram
    Apr 24, 2015 at 15:31
  • $\begingroup$ Yes but if we can make any assumptions here, then the question becomes trival to solve. It is not a well worded question. It should have stated restrictions and allowances very clearly. $\endgroup$
    – David
    Apr 24, 2015 at 15:34

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