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Solve 17d mod 24 = 1. Would it be d = 17 inverse mod 24 and then solved using EEA?

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closed as off-topic by Shailesh, Michael Hoppe, Lord Shark the Unknown, Leucippus, user10354138 Nov 13 '18 at 6:44

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Yes that means $$17d = 24k +1$$ or as you put it , $d$ is the multiplicative inverse of $17$ mod $(24).$

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Yes, you have it correct. Since $17d\equiv 1\;(mod\;24)$, there exists an integer $k$ such that $17d-1=24k$. This is a linear Diophantine equation, and since $gcd(17,24)=1$, there is a guarenteed solution. In this particular case, we have a solution of $d=17$ and $k=12$, which can be found using the Euclidean Algorithm.

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the Extended Euclidean algorithm applied to $17$ and $24$ will give us $k,m \in \mathbb{Z}$ such that

$$17k + 24m = \gcd(17,24)=1$$

and taking this equation modulo $24$ we get

$$17k \equiv 1 \pmod{24}$$

showing that $k$ is indeed the inverse of $17$ modulo $24$ (and $17$ of $k$ as well).

I get $5 \cdot 24 + (-7)\cdot 17 = 1$ with EEA.

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