Bribe The Friend Again!

We want the same amount numbers to be remaining after paying two pounds regardless of whether Kostya says yes once or no twice. Thus, we want:

$x=(1-x)^2$ , which gives $x=\frac{3-\sqrt{5}}{2}$

We can then see that $1-x=\frac{\sqrt{5}-1}{2}$ , the inverse of the golden ratio.

Since $144$ , is a number in the Fibonacci sequence, we see that we want to split the numbers into the Fibonacci sequence.

We split $144$ into $89$ and $55$ , getting the answer yes for the set of $55$ and no for the answer of $89$ . We can see that $1$ pound is paid per level of the Fibonacci sequence.

We can use this to go all the way down to $F_3=2$ , which will cost us $9$ pounds, since $144=F_{12}$ . Now, there are two numbers left. We have no way of knowing whether the answer will be yes or no, so we will need to pay $2$ pounds for it, because we are assuming the worst case scenario occurs.

$\therefore$ We put pay $\boxed{11}$ pounds in order to ensure that we can be certain of the number.