Question

One day a king left his castle with a bag of silver coins to wander his kingdom. To the first peasant he met, he gave half his coins plus two more. A little later, he met another peasant to whom he also gave half his coins plus two more. Walking on, he met a third peasant and again gave half his coins plus two more. Finally, the king went home with two coins left in his bag.

(a) How many coins did he have to begin with?

(b) What if he was left with "n" amount of coins?

(c) What if he gave away any unit fraction (1/a) each time?

(d) What if he gave away any unit fraction (1/a) each time and was left with any amount of coins (d)?

Answer 1

Explanation:

Let us assume the king before leaving his castle had x silver coins.

For the first peasant he met he gave

`(x)/(2) +2`

Now he is left with

`x-((x)/(2) +2)\\=(x)/(2) -2`

Next peasant he gave half of what he had and two more

Hence next peasant got

`(1)/(2) ((x)/(2) -2)\\=(x)/(4) -1`

Balance left =
`(x)/(2) -2-( (x)/(4) -1)\\=(x)/(4) -1`

Third peasant got half + 2 more

Hence third peasant got

`(1)/(2)( (x)/(4) -1)\\=(x)/(8) -(1)/(2)`

Balance he had

`(x)/(4) -1-((x)/(8) -(1)/(2) )\\(x)/(8) -(1)/(2)=2\\(x)/(8)=(5)/(2)\\x=20`