Since $(256,337)=1$, by Lemma 12 we know that $256 x \equiv$ $179(\bmod 337)$ has integer solutions. Since $337=256+81,256=81 \times$ $3+13,81=13 \times 6+3,13=4 \times 3+1$, we get $1=13$ $4 \times 3=13-4 \times(81-13 \times 6)=25 \times 13-4 \times 81=$ $25 \times(256-81 \times 3)-4 \times 81=25 \times 256-79 \times 81 \Rightarrow$ $25 \times 256-79 \times(337-256)=104 \times 256-79 \times 337$. Therefore, we have
$$104 \times 256 \equiv 1(\bmod 337)$$

By $256 x \equiv 179(\bmod 337)$ and Lemma 5, we have
$$104 \times 256 x \equiv 104 \times 179 \quad(\bmod 337)$$

Since $104 \times 179=55 \times 337+81$, we get
$$104 \times 179 \equiv 81(\bmod 337)$$

From (32) to (34), we get $x \equiv 81(\bmod 337)$.