Puzzles/Arithmetical puzzles/Eggs to Market/Solution

He started with 7 eggs. How? See Below: At the market the farmer had 1 egg. Working backward from toll gate two, the only way that could happen is by giving up half his eggs plus half an egg (dividing 3 into half to get 1.5 plus .5 to give up 2 eggs, leaving 1 egg). Again working backward from toll gate one, the only way that could happen is by giving up half his eggs plus half an egg (dividing 7 into half to get 3.5 plus .5 to give up 4 eggs, leaving 3 eggs). So, the farmer left the farm with 7 eggs! (this proof is contributed by Pankaj Kumar) Mathematical proof: To calculate exact value of eggs, lets back-track; At market, farmer has 1 egg. Lets assume, farmer has x eggs before passing Toll-Gate2 (before paying half eggs and an half egg) Total eggs paid at Toll-Gate2 will be : gate2 = ( (x/2) + (1/2) ) After passing, he is left with 1 egg . Which gives, ( x - gate2 ) = 1 Replacing gate2 x - ( (x/2) + (1/2) ) = 1 x - ( (x+1) / 2 ) = 1 ( 2x - x - 1 ) / 2 = 1 x - 1 = 2 x = 3 So, before passing Toll-Gate2 farmer has 3 eggs. Now, assume before passing Toll-Gate1 farmer has y eggs ( at initial stage he as this many eggs) Total eggs paid at Toll-Gate1 will be : gate1 = ( (y/2) + (1/2) ) After passing, he is left with 3 egg . Which gives, ( y - gate1 ) = 3 Replacing gate1 y - ( (y/2) + (1/2) ) = 3 y - ( (y+1) / 2 ) = 3 ( 2y - y - 1 ) / 2 = 3 y - 1 = 6 y = 7 So, the farmer left the farm with 7 eggs!