There are 5 bags labeled 1 to 5. All the coins in a given bag have the same weight. Some bags have coins of weight 10 gm, others have coins of weight 11 gm. I pick 1, 2, 4, 8, 16 coins respectively from bags 1 to 5. Their total weight comes out to 323 gm. Identify the bags with 11 coins.
I can solve the problem by using the fact that sum of odd number and even number is odd number
$$x_1+2x_2+4x_3+8x_4+16x_5=323$$ $$odd+even=odd$$ hence $$x1=11$$ $$2x_2+4x_3+8x_4+16x_5=323-11=312$$ dividing by 2
$$x_2+2x_3+4x_4+8x_5=312/2=156$$ $$even + even=even$$ hence $$x2=10$$ similarly continuing ,will get all weights of coin
I found this method very randomly.
I don't have logical reason for dividing by 2 step after each iteration?
is there any better method? what is the reasoning behind this method? is there any recursion happening? how can we prove that this method work?
First find out how many of the 31 coins (1 + 2 + 4 + 8 + 16) weigh 11 gm. You will find there are 13 of them. Then think of how you can write 13 as a sum of powers of 2 - this is "writing 13 in base 2". Answer, 8 + 4 + 1. So the coins that came from bags 1, 2 and 4 weigh 11 gm each, and the others 10 gm each.