Given that $33^{-1} \equiv 77$ (mod $508$), find $11^{-1}$ (mod $508$) as a residue modulo 508. (Give an answer between 0 and 507, inclusive.)
+287 By definition $33^{-1} \equiv 77 \pmod{508}$ means $33\cdot 77 \equiv 1 \pmod{508}$. Thus $11\cdot 3\cdot 77 \equiv 1 \pmod{508}$. Hence, $11^{-1} \equiv 3\cdot 77 \equiv 231 \pmod{508}$.
+26387 Given that \(33^{-1} \equiv 77 \pmod{508}\),
find \(11^{-1}\pmod{508}\) as a residue modulo \(508\).
(Give an answer between 0 and 507, inclusive.)
\(\begin{array}{|rcll|} \hline 33^{-1}= \dfrac{1}{33} & \equiv& 77 \pmod{508} \\\\ \dfrac{1}{33} = \dfrac{1}{3*11} & \equiv& 77 \pmod{508} \\\\ \dfrac{1}{3*11} & \equiv& 77 \pmod{508} \quad | \quad * 3 \\\\ \dfrac{3}{3*11} & \equiv& 3*77 \pmod{508} \\\\ \dfrac{1}{11} & \equiv & 231 \pmod{508} \\\\ \dfrac{1}{11}=\mathbf{ 11^{-1} } & \equiv & \mathbf{ 231 \pmod{508} } \\ \hline \end{array}\)
