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 $x \equiv 3515 \pmod{5824}$. 

$x=32+81k\equiv 32+17k \equiv 59 \pmod {64}$
$k\equiv \frac{27}{17}\equiv\frac{27}{81}\equiv \frac{1}{3}\equiv \frac{129}{3}\equiv43 \pmod {64}$
$x=32+81(43+64k')=3515+5184k'$
