If $a,b,x,$ and $y$ be real numbers with $a>4$ and  $b>1$such that $\frac{x^2}{a^2}+\frac{y^2}{a^2-16}=\frac{{(x-20)}^2}{b^2-1}+\frac{{(y-11)}^2}{b^2}=1.$ then the least possible value of $\left[\frac{a+b}{5}\right]$ is equal to______ (where [.] denotes greatest integer function).

Denote  $P=(x,y).$
Because $\frac{x^2}{a^2}+\frac{y^2}{a^2-16}=1,P$ is on an ellipse whose center is $(0,0)$ and foci are  $(-4,0)$and $(4,0)$.
Hence, the sum of distance from P to$(-4,0)$ and $(4,0)$ is equal to twice the major axis of this ellipse 2a.
Because $\frac{{(x-20)}^2}{b^2-1}+\frac{{(y-11)}^2}{b^2}=1,P$ is on an ellipse whose center is $(20,11)$and foci are$(20,10)$ and $(20,12)$.
Hence, the sum of distance from P to$(20,10)$  and$(20,12)$ is equal to twice the major axis of this ellipse 2b.
Therefore $2a+2b$,is the sum of the distance from P to four foci of these two ellipses.
To make this minimized, P is the intersection point of the line that passes through$(-4,0)$ and $(20,10)$ , and the line that passes through $(4,0)$and  $(20,12).$
The distance between $(-4,0)$ and $(20,10)$ is  $\sqrt{{(20+4)}^2+{(10-0)}^2}=26.$
The distance between $(4,0)$ and $(20,12)$ is  $\sqrt{{(20-4)}^2+{(12-0)}^2}=20.$
Hence, least value of  $2a+2b=26+20=46.$
Therefore, least value of  $a+b=23$