Bézout identity for 1001 and 315

Euclidean algorithm.
\begin{align*}
1001 &= 3\cdot315+56,\\
315 &= 5\cdot56+35,\\
56 &= 1\cdot35+21,\\
35 &= 1\cdot21+14,\\
21 &= 1\cdot14+7,\\
14 &= 2\cdot7+0.
\end{align*}
So $\divisibility(1001,315)=7$.

Back-substitution.
\begin{align*}
7 &= 21-14\\
&=21-(35-21)=2\cdot21-35\\
&=2(56-35)-35=2\cdot56-3\cdot35\\
&=2\cdot56-3(315-5\cdot56)=17\cdot56-3\cdot315\\
&=17(1001-3\cdot315)-3\cdot315\\
&=17\cdot1001-54\cdot315.
\end{align*}
So $17\cdot1001-54\cdot315=7$.