Solution to Puzzle #128: The Milkman Again

Apologies for not posting last weekend, I was traveling, and am likely to be traveling quite a bit in the next month, so there will be some irregularity in posting.

This was a relatively simple one, lots of people sent correct answers that included Pratik Poddar, Aditya GV, P R Kumar, Suman Saraf, Arjun Raychaudhuri, Shruti Mittal, Girish Tutakne, Anubhav Garg and Abhishek Masand. Well done all!

There is clearly more than one way to do it, I am reproducing the one from Girish Tutakne.

Let us name the jars A (for 24 lit), B (13 lit), C(11 Lit) and D(5 lit). I will use the nomenclature for example A(10) to represent that jar A has 10 lit of milk.
Initial condition: A(24); B(0); C(0); D(0)
Step 1: A(8); B(0); C(11); D(5) —-> poured milk from jar A to fill up jars C and D
Step 2: A(8); B(13); C(0); D(3) —-> poured milk first from jar C completely and then remainder from jar D to fill up jar B
Step 3: A(8); B(13); C(3); D(0) —-> poured milk from jar D to jar C
Step 4: A(8); B(8); C(3); D(5) —-> poured mils from jar B to fill up jar D
Step 5: A(8); B(8); C(8); D(0) —-> poured milk from jar D to jar C. We now have 3 jars with 8 lit each

Hope you all enjoyed the puzzle!


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