How do you encrypt $51$ with public key $(n,e) = (91,23)$
I understand that $c = 51^{23} \bmod 91$. How can I calculate the result on a calculator?
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Sign up to join this communityHow do you encrypt $51$ with public key $(n,e) = (91,23)$
I understand that $c = 51^{23} \bmod 91$. How can I calculate the result on a calculator?
If your calculator is able to compute $n^2$, you can compute $m^e \bmod n$ using the binary exponential method.
In this method, you should first compute the binary form of $e$. Let $\ell$ be the number of bits in $e$, and let $e_i$ denote the $i$-th bit of $e$, so that $e=\sum\limits_{i=0}^\ell e_i \cdot 2^i$.
Now, with the algorithm below, you can compute $c$:
$z:=1$
$\text{for } i:= \ell \text{ down to } 0 \text{ do:}$
$\quad z:=z^2 \bmod n$
$\quad \text{if } e_i = 1 \text{ then } z:=(z \cdot m) \bmod n$
$\text{end for}$
$\text{return } z$