<s>[INST] Solve $x \equiv 32 \pmod{81}$ and $x \equiv 59 \pmod{64}$. 
Solve $x \equiv 32 \pmod{81}$ and $x \equiv 59 \pmod{64}$. 

$32 + 81k = 59 + 64n \implies 81k - 64n = 27$
$17k \equiv 27 \pmod{64}$.
$64 = 3(17) + 13$
$17 = 1(13) + 4$
$13 = 3(4) + 1$ 
So $1 = 13 - 3(4) = 13 - 3[17 - 13] = 4(13) - 3(17) = 4(64 - 3*17) - 3*17 = 
4(64) - 15(17)$. Thus $k \equiv 1 \pmod{64} \implies k = 1 + 64y$ so we have $n = \frac{54 + 5184y}{64}$
But this is not possible . Help?
EDIT $k\equiv 43 \pmod{64}$ thus,
$k = 43 + 64y$ thus $x = 32 + 81(43 + 64y) = 3515 + 5824y$ so then $x \equiv 3515 \pmod{5824}$. 
 [/INST] $x\equiv 32 (\mod 81) \  and  \ x\equiv 59 (\mod 64) \Rightarrow  x=81a+32 \ and \ x=64b+59 \\ $
then $81a+32=64b+59 \\
81a-64b=27 \ ...I \\$
with the euclidean algorithm we get a solution 
$ (a,b)=(-405,-513) \\ 
81(-405)-64(-513)=27 \ ...II \\ $
subtracting II from I we get $ 81(a+405)-64(b+513)=0 \Leftrightarrow 81(a+405)=64(b+513) \ ...III \\ $
$64\mid a+405\Rightarrow \boxed{a=64k-405} \\ $
plugging $a$ into III gives $\boxed{b=81k-513} \\$
plugging $a$ into $x=81a+32$ or $b$ into $x=64b+59\\$
$\boxed{x=5184k-32773} \ \ k\in \mathbb{Z}$
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