King Lear and his three daughters

Working Backwards is the Key!

0+1= 1; 1 2= 2; 2+1= 3; 3+1= 4; 4 2= 8; 8+1= 9; 9+1= 10; 10*2= 20; 20+1= 21

Bravo, 21.

What if the King had $n$ daughters?

Let $c_n$ be the least number of coins he would have to bring to keep $n$ daughters (and their guards) happy. To leave a daughter with $c$ coins in his royal pockets, he had to arrive with $2c+3$ . Thus $c_{n}=2c_{n-1}+3$ or $c_{n}+3=2(c_{n-1}+3)$ , so that $c_{n}+3=2^{n}(c_{0}+3)=3\times2^n$ and $\boxed{c_{n}=3(2^n-1)}$ .

For his three daughters he has to bring $3\times7=\boxed{21}$ coins.

Let the number of coins the king had be $x$ then :

When he visits Regan he gives 1 coin to gaurd, now he is left with $x-1$ coins,

Then he gives half to Regan and now is left with $\frac{(x-1)}{2}$ coins

Now he gives one to the gaurd again leaving him with $\frac{x-3}{2}$ coins.

Similarly, now he goes to visit Generil, gives one to gaurd and is left with $\frac{x-5}{2}$

Then he gives half to Generil and has $\frac{x-5}{4}$ coins left

Before leaving he gives a coin to the gaurd, now he has $\frac{x-9}{4}$ coins left.

Same goes with Coedelia, the King gives a coin to gaurd and is left with $\frac{x- 13}{4}$

Now he gives half coins to Cordelia, so coins left = $\frac{x-13}{8}$

Then gives a coin to the guard, remaining coins = $\frac{x-21}{8}$

Now, since he has 0 coins left we equate $\frac{x-21}{8}$ to 0, therefore: $\frac{x-21}{8} = 0$

$x-21=0$

$x=21$

Let the number of gold coins to start with be $n$ , and the numbers Regan, Goneril and Cordelia get be $r$ , $g$ and $c$ respectively.

Then $n = 1+r+1+1+g+1+1+c+1 = r+g+c+6$

We note that:

$\begin{aligned} r & = \dfrac {n-1}{2} \\ g & = \dfrac {1}{2} \left( n - \dfrac {n-1}{2} - 3 \right) = \dfrac {1}{2} \left( \dfrac {2n - n+1-6}{2} \right) = \dfrac {n-5}{4} \\ c & = \dfrac {1}{2} \left( n - \dfrac {n-1}{2} - \dfrac {n-5}{4} - 5 \right) \\ & = \dfrac {1}{2} \left( \dfrac {4n - 2n+2-n+5-20}{4} \right) = \dfrac {n-13}{8} \\ \Rightarrow n & = \dfrac {n-1}{2} + \dfrac {n-5}{4} + \dfrac {n-13}{8} + 6 \\ & = \dfrac{4n-2+2n-10+n-13+48}{8} \\ \Rightarrow 8n & = 7n + 21 \\ \Rightarrow n & = \boxed{21} \end{aligned}$