Let $x$ be a real number such that $\sin^{10}x+\cos^{10} x = \tfrac{11}{36}$. Then $\sin^{12}x+\cos^{12} x = \tfrac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m+n$.
We can substitute $y = \sin^2{x}$. Since we know that $\cos^2{x}=1-\sin^2{x}$, we can do some simplification. 
This yields $y^5+(1-y)^5=\frac{11}{36}$. From this, we can substitute again to get some cancellation through binomials. If we let $z=\frac{1}{2}-y$, we can simplify the equation to:
\[\left(\frac{1}{2}+z\right)^5+\left(\frac{1}{2}-z\right)^5=\frac{11}{36}.\] 
After using binomial theorem, this simplifies to:
\[\frac{1}{16}(80z^4+40z^2+1)=\frac{11}{36}.\]
If we use the quadratic formula, we obtain  $z^2=\frac{1}{12}$, so $z=\pm\frac{1}{2\sqrt{3}}$ (observe that either choice of $z$ doesn't matter). Substituting $z,$ we get:
\[\sin^{12}{x}+\cos^{12}{x}=\left(\frac{1}{2}-z\right)^6+\left(\frac{1}{2}+z\right)^6=2z^6 + \frac{15z^4}{2} + \frac{15z^2}{8} + \frac{1}{32}=\frac{13}{54}.\]
Therefore, the answer is $13+54=\boxed{067}$.
-eric2020, inspired by Tommy2002

Motivation
The motivation to substitute $z=\frac{1}{2}-y$ comes so that after applying the binomial theorem to $y^5+(1-y)^5=\left(\frac{1}{2}+z\right)^5+\left(\frac{1}{2}-z\right)^5,$ a lot of terms will cancel out. Note that all the terms with odd exponents in $\left(\frac{1}{2}+z\right)^5$ will cancel out, while the terms with even exponents will be doubled. 
mathboy282