<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] $$x = 59 + 64k$$
$$x = 32 + 81j$$
$\text{lcm}(81, 64) = 5184$
so
$$81x = 4779 + 5184k$$
$$64x = 2048 + 5184j$$
Subtract equations
$$17x = 2731 + 5184(k-j)$$
or
$$17x \equiv 2731 \bmod 5184$$
$\gcd(17, 5184)=1$, so:
$$x \equiv 2731 \cdot 17^{-1} \bmod 5184$$
$$x \equiv 2731 \cdot 305 \equiv 832955 \equiv 3515 \bmod 5184$$
Answer is
$$x \equiv 3515 \bmod 5184$$
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