# 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!