I have 3 cans of sizes 7, 9 and 16 liters resp. What is the minimum number of operations required to deliver 1 liter? I was reading about elementary number theory when I came across this question. I've seen such questions before but never found any systemic approach on how to solve them. Does there exist one? Please elaborate since I'm just a beginner.
 A: Considering the reference given  by Jean Marie in comment, for a mathematical model let x, y and z be the numbers of filing of 16, 9 and 7 liter cans respectively. If x or y or z is positive it means you fill the can, if each of them is negative it means the related can is filled by another can. Now we construct a linear Diophantine equation with three unknown x, y and z as below:
$16x+9y+7z=1$
Thus equation gives the number of filling of each can from the tank (filling) and transitions(from a can to another can). $1$ on the RHS is the result of algebraic sum of terms and gives the volume of liquid finally remains in a can which you want to be one liter. Meanwhile you want the sum of operation $S=x+y+z$ be minimum. We solve above equations and take small values :
$16x+9y+7z=1$
$(x, y, z)=(5, -8, -1),(6, -9, -2), (-2, -1, 6), (-1, -2, 5), (-1, 5, -4), (-2, 6, -3)$
The sum of operation is:
$S=|x|+|y|+|z|$
So the related sums of above results are:
$S=14, 17, 9, 8, 10, 11$
Hence minimum operations is $8$ which results from $(x, y, z)=(-1, -2, 5)$
Which suggests following operations:
Five times filling the 7 Liter can from the tank. One time Pouring  in 16 Liter can and two times  in 9 Liter can to fill them by filled 7 Liter can ; what finally remains in 7 Liter can is:
$5\times 7-1\times 16-2\times 9=1$
