Jar to measure 6 liters

I have two Jars 5 liters and 7 liters and a well to take water.
I want to measure exact 6 liters with out using a 3rd jar.
Please help
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PeteEngineerAsked:
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lazyberezovskyCommented:
Here your jars and steps :)
               5liters                                   7liters
1)    <empty>                                      fill 7 liters
2)   get 5 liters from second jar       2 liters left
3)   empty this jar                                2 liters
4)   get 2 liters from second jar           <empty>
5)  2 liters                                         fill 7 liters
6)  get 3 liters from second jar           4 liters left
7) empty this jar
8) get 4 liters from second jar            <empty>
9) 4 liters                                           fill 7 leters
10) get 1 liter from second jar           6 liters left BINGO!
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sdstuberCommented:
6) get 3 liters from second jar           4 liters left


how are you measuring 3 liters?
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phoffricCommented:
If you have two jars having an integral number of liters, k and m, respectively, then if k and m are mutually prime (i.e., they share no prime factor other than 1), then you can ask for any amount of liters <= k. It may take a hundred steps (or a thousand) but it is doable.

When considering gcd, we know that for mutually prime numbers, k and m, gcd(k,m) = 1, and so there exists integers a, b such that ak + bm = 1.
This is the basis solving the above problem.
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phoffricCommented:
>> how are you measuring 3 liters?
Looks like he poured the 7 liters into the 5 liter jar (which already had 2 liters), and he filled the 5 liters to capacity; so 3 liters was removed from the 7 liters leaving only 4 liters in the 7 liter jar.
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lazyberezovskyCommented:
sdstuber,

I get it by filling first jar. It has volume 5liters. And 2 already in. So, only 3 liters could be added more.
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sdstuberCommented:
gotcha,  thanks
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phoffricCommented:
So, if I had a 27 liter jar (3^3) and a 64 liter jar, I should be able to measure any number of liters from 1 to 64.
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Infinity08Commented:
only consuming 20 liters total this way (instead of 21 for lazyberezovsky's approach) :

      5 liters            7 liters
1)      fill 5 liters            empty
2)      empty                  get 5 liters from first jar
3)      fill 5 liters            5 liters
4)      3 liters left            get 2 liters from first jar
5)      3 liters            empty this jar
6)      empty                  get 3 liters from first jar
7)      fill 5 liters            3 liters
8)      1 liter left            get 4 liters from first jar
9)      1 liter                  empty this jar
10)      empty                  get 1 liter from first jar
11)      fill 5 liters            1 liter
12)      empty                  get 5 liters from first jar
13)      empty                  6 liters

I'm all for preserving this precious resource lol ;)

Merely for fun - don't award points please.
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phoffricCommented:
Yes, there are two ways to solve the problem. One approach is to start off with the smaller jar being empty, and the other approach has the larger jar being empty. I don't know how to predict which approach reaches the goal faster. (I think I used to know.)
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jkunreinCommented:
Just to add to this fun little exercise...

You can solve these types of problems with a Diophantine equation.  For this problem, you write it as:
5x + 7y = 6

You assign the two known jars as coefficients of x and y, and you are setting it equal to the final value you want.  Any integer solution of x and y will tell you how many times you need to fill one jar and how many times you dump the other jar.

For example, you can see that (-3, 3) solves the equation. The -3 means that you dump out the x jar (in this case, 5 liters) three times and that you fill the y jar (7 liters) three times.  If you follow lazyberezovsky's solution, you see that's exactly what happened.  Well, the third dumping of the 5-liter jar was understood.  You actually have 11 liters between the two jars, but the dumping of the 5 liters is a given once you obtain your goal.

Also interesting is that there is more than one way to do this.  Notice that (4, -2) is another solution.  This means that you can get your 6 liters by filling up the 5-liter jar four times while dumping the 7-liter jar twice.  

And, as phoffric points out, this works best if the two jar values are relatively prime, but it also depends on whether the common factor is also a factor of the final value.  You can see more about it at: http://everything2.com/title/Two+water+jugs+problem

But, in short, you can write these problems out as ax + by = c.  If there is an integer solution for x and y then this can be solved. Technically, there are an infinite number of solutions, but most of them would just repeat the two most basic solutions.

Fun times.
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