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Calculate $181^{-1} (mod\,29)$. As part of the calculation write $181 * A ≡ B (mod\,29)$ where $A \& B$ between 0 and 28.

These are my calculations:

$$181 (mod\,29) ≡ 7(mod\,29)$$

$$=>181^{-1}(mod\,29) ≡ 7^{-1} (mod\,29)$$

Because 29 is a prime number, so with fermat's theorem: $7^{28} (mod\,29) ≡ 1$

$$7^{-1} (mod\,29) ≡ 7^{27} (mod\,29) ≡ 7 * 7^{26} (mod\,29) ≡ 181 * 7^{26} (mod\,29)$$

$$A = 7^{26} (mod\,29) = 16$$

$$B = 7^{27} (mod\,29) = 25$$

It was incorrect so I'm hoping someone points out my mistake.

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  • $\begingroup$ Umm.... you never actually calculated what $181^{-1}$ was. ANd you didn't use it to make you equation. But if you had and $181^{-1} = 25$ then you could chose arbitrary $B$ and solve for $A = 181^{-1}B = 25 B$ $\endgroup$
    – fleablood
    Aug 27, 2020 at 7:59
  • $\begingroup$ $181^{-1}\equiv 7^{27}\equiv 25$ is correct, but there are simpler ways to compute it (see the linked dupes). This means $\,181\cdot 25\equiv 1\,$ so we can choose $\,A\equiv 25,\ B\equiv 1\ $ (rather than scale them by $7^{-1}$ as you essentially did). $\endgroup$ Aug 27, 2020 at 14:16

2 Answers 2

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We have $181\equiv7\mod{29}$, as you said. We know $4\cdot7=28\equiv-1\mod{29}$, so $7(-4)\equiv1\mod{29}$ or $$181\cdot25\equiv1\mod{29}$$ and $$181^{-1}\equiv25\mod{29}$$

I don't really understand your calculation, so I can't really point out your mistake. In particular, I don't see where you ever write some thing in the form $181A\equiv B\mod{29}$ with $A$ and $B$ between $0$ and $28$. You are correct that the answer is $7^{27}$ but I don't know how you reduced that to $25$ modulo $29$.

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$181^{-1} \equiv 7^{-1}$ but what is that. You could do $7^{-1}\equiv 7^{27}$ and do successive squaring but that's a pain.

We need $7x \equiv 1 \pmod 29$. Let's pretend I don't see $7*4 \equiv -1$ so $7*(-4)\equiv 1$ and instead do that there is an integer $k$ so that $7x = 29k + 1$ so $x = 4k+ \frac{k+1}7\in \mathbb Z$ so $k\equiv -1 \pmod 7$ say if $k = -1$ then $x \equiv -4\equiv 25 \pmod {29}$

Now we use that to come up with examples where $181*A = B$ means $A\equiv 181^{-1}*181 A\equiv 181^{-1}*B \equiv 25B$ so if, say $B=2$ then $A\equiv 50\equiv 21$, for example.

Actually, as a problem being asked of you, this was a pretty stupid one. You could have found an equation just as easily by arbitrarily choosing an $A$, say $A=2$ and letting $B\equiv 181A \equiv 7A \equiv 7*2\equiv 14$ so $B = 14$.

I wouldn't have marked yours incorrect. But you never actually stated $181^{-1}\equiv 25$ and you didn't use that to calculate $A,B$.

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