Dragon Curses

The Dragons

Each of three villages in StackVille is being tormented by a robotic AI dragon living in a cave above the town. The villages are Arquade, Ubutu, and Tezos.



The town's Robotic Artificial Intelligence Programmers BrynSen , MykRowe, and KitNyck created these three dragons [ yep, they had AI in StackVille way back then ]. Initially they were made for defending StackVille, but the project was killed by government leaders. "Too costly", the bean counters cried.

These programmers were mad they did not receive any recognition or gratitude for their recent work in AI development, so in retaliation, they created Robotic Dragons to torment and curse, the villages.

The Dragon's Peudocode

MykRowe's dragon is programmed to menace (curse) one of the villages for the same number of years as the square root of the number of years of KitNyck's dragon curse on a second village.

MykRowe's dragon curse will also last the number of years equal to half the square root of the number of years of BrynSen's dragon curse on the third village.

KitNyck's dragon curse will last the number of years equal to twice the square root of the number of years of BrynSen's curse.

How long will each curse last?

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2 Answers 2


Let $b$, $m$, and $k$ be the numbers of years for the three dragons BrynSen, MykRowe, and KitNyck.

Then we have three equations:

$$m = \sqrt{k},\quad\quad m = \frac{1}{2}\sqrt{b},\quad\quad k = 2\sqrt{b},$$


$$k=m^2,\quad\quad b=4m^2,\quad\quad k^2=4b$$.


$k^2=4b=16m^2$ which means $m^4=16m^2$.


$m=4$, $k=16$, $b=64$.


Here's my answer:

MykRowe's Dragon's curse lasts $4$ years, BrySen's Dragon's curse last $64$ years, and KitNyck's Dragon's curse lasts $16$ years.


Let's say MykRowe's Dragon=$M$, BrynSen's dragon=$B$, and KitNyck's dragon=$K$. The three statements gives us the following three equations:$$\begin{align}(1)M&=\sqrt{K}\\(2)M&=\frac{\sqrt{B}}{2}\\(3)K&=2\sqrt{B}\end{align}$$Plugging in $(1)$ into $(2)$ yields $\sqrt{K}=\frac{\sqrt{B}}{2}\implies K=\frac{B}{4}$. Plugging this into $(3)$ gives $\frac{B}{4}=2\sqrt{B}$. Solving gives us $B=\boxed{64}$. Plugging this into $(3)$ gives $K=2\sqrt{64}=\boxed{16}$. Plugging this into $(1)$ gives $M=\sqrt{16}=\boxed {4}$.


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