# Tagged Questions

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### Residue class of a huge repunit modulus a huge number

Given a number with only 1: X = 1111...1 (N times 1 in total), and another number M, I want ...
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### How to find out a^b^c^… mod m

I would like to calculate: abcd... mod m I know when a is coprime to m then we can easily find out the answer using Euler's totient function. But I wish to know the ideas when a is not coprime to ...
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### Find modular inverse of a number

Recently I have read extended euclid's algorithm which is used to find out the modular inverse of a number N whith respect to MOD such that $\gcd(N,MOD)=1.$ But I have a doubt that how to find modular ...
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### Is it possible to do modulo of a fraction

I am trying to figure out how to take the modulo of a fraction. For example: 1/2 mod 3. When I type it in google calculator I get 1/2. Can anyone explain to me how to do the calculation?
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### PowerMod: Solving for the base

Given the problem $c^d \mod n = m$ and values for $d$, $n$, and $m$, how would one solve for $c$? A general solution or approach would be fine, as well as the values for my specific problem are as ...
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### Solving Equations in $\mathbf{Z}/n\mathbf{Z}$ with Indices

Consider the equation $x^4 = 7,$ which we wish to solve in $\mathbf{Z}/29\mathbf{Z}.$ I was taught a technique for solving this problem, but I can't understand it. I'll try my best to describe it, ...
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### calculate reverse number with 2 conditions

I can't find the reversed number of $2 \mod 13$ ($2^{-1}=?$) that is also a solution to $$5x = 2 \mod 13.$$ How can I find it? Thanks!
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### Number theory notation

I am confused with the below notations . I know that ($a \equiv b \mod {n} )\iff ( n|(a-b)$ ) but what the below notation says ? $a = b \mod {n}$ and in theorem 16 in this ,it's given as below ...
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### Number theory problem.Primes modules.

If $$a^p\equiv b^p \pmod p$$ where $p$ is prime prove that $$a^p\equiv b^p \pmod{p^2}$$ that problem was at my exam today on number theory and i just didnt have a clear mind to solve it.Although i ...
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### Correct reasoning when proving the multiplication property in modular arithmetic?

I am trying to understand why this rule works: \begin{align*} a \equiv b \pmod c \quad k \equiv j \pmod c \qquad &\implies \qquad ka\equiv jb \pmod c \end{align*} I saw that the proof is ...