Measure 11 liters using bottles of 16, 6, and 3 liters 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:


*

*We cannot measure a bottle half-way.

*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)

*There's no reservoir where we can keep the spilled water.

*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.
 A: 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.
A: With a bit of lateral thinking, this is possible like this:


*

*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.

*Now 8 liters remain in the 16 liter bottle, so just add the 3 liters.

A: If you have an extra reservoir that can contain at least 16 liters, it is possible:


*

*Start with 16-6-3-0

*Empty bottles of 6 and 3 into reservoir: 16-0-0-9

*Fill bottle of 6 from bottle of 16: 10-6-0-9

*Empty bottle of 6 into reservoir: 10-0-0-15

*Fill bottle of 6 with bottle of 16: 4-6-0-15

*Empty bottle of 6 into reservoir: 4-0-0-21 

*Move content of bottle of 16 to bottle of 6: 0-4-0-21

*Refill bottle of 16 from reservoir: 16-4-0-5

*Fill bottle of 3 from bottle of 16: 13-4-3-5

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


*

*empty a bottle completely; and 

*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):

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

A: 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...).
A: 


*

*Discard all the water from $3$ liter bottle

*Fill it full with water from $16$ liter bottle

*Now, $16$ liter bottle has $13$ liter water, mark the level of the water

*Discard all the water from $3$ liter bottle

*Fill it full with water from $16$ liter bottle

*Now, $16$ liter bottle has $10$ liter water, mark the level of the water

*Divide the portion of the bottle between $10$ and $13$ mark into $3$ equal parts

*Fill one part with water from any one of the other two bottles

