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Assume we are working in a cyclic group $G$ with generator $g$ and let $A$ denote the public key in use. From the definition of ElGamal encryption, we have $R_i = g^{r_i}$ and $c_i=A^{r_i}\cdot m_i$, where $r_i$ is some random number, for $i\in\{1,2\}$. Hence, with $R:=R_1\cdot R_2$ and $c:=c_1\cdot c_2$ (where $\cdot$ denotes $G$'s operation), we have ...

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If I'm understanding your question right, you want to obtain $d$ from given $n$ and $e$. You'll have to factor $n=33=3*11$ and as $N=p*q$ you have obtained your $p=11$ and $q=3$. Now proceed as usual with calculating the inverse. As pointed out correctly above, you can't easily generalize this approach to larger numbers as factoring $n$ will be infeasible. ...

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The standard algorithm used for RSA encryption and decryption is exponentiation by squaring. The basic idea is to write the exponent out in binary. For example, for $d = 4267793$, \begin{aligned} 4267793 &= 10000010001111100010001_2 \\ &= 2^{22} + 2^{16} + 2^{12} + 2^{11} + 2^{10} + 2^9 + 2^8 + 2^4 + 2^0.\end{aligned} Now, given some RSA ...

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