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The table below shows the multiplicative inverses of the first $9$ positive residues modulo $47$.

\begin{tabular}{c|ccccccccc}
$b$ & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ \hline
inverse of $\,b$ & 1 & 24 & 16 & 12 & 19 & 8 & 27 & 6 & 21
\end{tabular}

Find the multiplicative inverse of $10\pmod{47}$.  Express your answer as an integer from $0$ to $46$, inclusive.

 Aug 22, 2023
 #1
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It's 43.

 Aug 23, 2023

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