# Multiplicative inverse of 97 modulo 386?

A little help would be a lifesaver :)

I already used the Euclidean algorithm to find the GCD of $386$ and $97$, which is $1$. However, I'm stuck on this question posed by my professor: "Then use your computation to explain the statement: $241$ is the multiplicative inverse of $97$ $\mathrm{modulo}$ $386$."

I've been stuck on the topic of multiplicative inverses for days so any help would be greatly appreciated. Thank you so much!

Using Extended Euclidean algorithm determine $p$ and $q$ such that $$97 \cdot p - 386\cdot q = 1$$ Then $97^{-1} \equiv p \mod 386$.

There is likely a typo in the question.

Using the Euclidean algorithm, we have that: \begin{align*} 386 &= 3(97) + 95 \\ 97 &= 1(95) + 2 \\ 95 &= 47(2) + 1 \end{align*} Working backwards, we see that: \begin{align*} 1 &= 95 - 47(2) = 95 - 47(97 - 95) \\ &= -47(97) + 48(95) = -47(97) + 48(386 - 3(97))\\ &= 48(386) - 191(97)\\ \end{align*} Hence, we see that: $$1 \equiv 48(386) - 191(97) \equiv -191(97) \equiv 195(97) \pmod{386}$$ So the multiplicative inverse of $97$ modulo $386$ is $195$.

Using formulae used here,

$$\frac{386}{97}=3+\frac{95}{97}=3+\frac1{\frac{97}{95}}$$

$$=3+\frac1{1+\frac2{95}}=3+\frac1{1+\frac1{\frac{95}2}}=3+\frac1{1+\frac1{47+\frac12}}$$

So, the previous convergent of $\displaystyle \frac{386}{97}$ is $\displaystyle3+\frac1{1+\frac1{47}}=3+\frac{47}{48}=\frac{191}{48}$

$$\implies386\cdot48-191\cdot97=1$$ $$\implies -191\cdot97=1\pmod{386}\implies 97^{-1}\equiv-191\equiv386-191$$