The Wrecked Rectangle

In order to scale it down, the lengths must be divided by $\sqrt { 10 }$ (or the area divided by 10).

This gives: $\frac { 5 }{ \sqrt { 10 } } \times \frac { 4 }{ \sqrt { 10 } } =2$ for the area.

Therefore: $2=\sqrt { x\sqrt { y\sqrt { x\sqrt { y\sqrt { x... } } } } }$

Notice the embedded equation and this can be reduced to the form:

$2=\sqrt { x\sqrt { y(2) } }$

$4=x\sqrt { 2y }$

$\frac { 4 }{ x } =\sqrt { 2y }$

$\frac { 16 }{ { x }^{ 2 } } =2y$

${ x }^{ 2 }y=8$

From this it can be worked out that the only values for x and y are 1,8,2 & 2 and therefore the answer is 13 [The factors of 8 are 1,2,4 & 8. With 1 and 4 both being square numbers these both fit into the form ${ x }^{ 2 }y=8$ ] =)

The area is $\frac{5}{\sqrt(10)} * \frac{4}{\sqrt(10)} = 2$

Let the expression be denoted by $z$ , therefore we have that:

$z = 2$

$\Rightarrow z^2 = x\sqrt{yz} = x\sqrt{2y} = 4$

$\Rightarrow 2x^2y = 16$

$\Rightarrow x^2y = 8 = 2x\sqrt{2y}$

$x^2y-2x\sqrt{2y} = 0 \Rightarrow x\left(xy - 2\sqrt{2y}\right) = 0$

$x \neq 0 \Rightarrow xy - 2\sqrt{2y} = 0$

$xy = 2\sqrt{2y} \Rightarrow x^2y^2 = 8y$

$x^2y^2 - 8y = 0 \Rightarrow y\left(x^2y-8\right) = 0$

$y \neq 0 \Rightarrow x^2y = 8, \{x,y\} \subset \mathbb{Z^+}$

Therefore $x^2$ must be a square factor of $8$ so $x^2 = 1, 4$

$x^2 = 1 \Rightarrow x=1, y=8$

$x^2 = 4 \Rightarrow x=2, y=2$

$1+8+2+2 = 13$

real crazy mathematician

Let r ( ) be square root:

Let z = [4/ r (10)] x [5/ r (10)] = 2;

(z^2/ x)^2/ y = z

=> z^3 = x^2 y

=> 8 = x^2 y

(x, y) = (1, 8) or (2, 2) only.

Check by substitution:

r (r (8 r (r (8 r (r (8 r (1))))))) = 1.978+ and

r (2 r (2 r (2 r (2 r (2 r (2 r (2))))))) = 1.978+

Answer = 1 + 8 + 2 + 2 = 13.