# RSA cryptosystem test question - help please!

Suppose that the 26 symbol alphabet $A,...,Z$ is used for all plaintext and ciphertext messages in an RSA cryptosystem. Suppose also that plaintext message units are length $2$ and ciphertext units are length $3$.

A user has a public key $(943,3)$

i) Encrypt for transmission to A the message

MATHEMATICAL

Okay, so $(943,3)$ = (n, e). $k=2, l=3$.

$943=23*41$

MATHEMATICAL = $12, 0, 19, 7, 4, 12, 0, 19, 8, 2, 0, 11$

What are the next steps? How do I finish this? The example is my notes is fairly complicated and uses Maple for one of the calculations which is confusing me as we have to do this in a class test? .. Without access to any computer programs...

If someone could explain this to me, that would be great! :)

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1st, $943\ne21\times41$. Second, one thing you have to do, before you can do any encrypting, is turn the message into a number (or string of numbers). You have to look at that example in the notes to see how they do that. This is before Maple or anything complicated gets done. –  Gerry Myerson Mar 1 '13 at 12:07
I meant 23*41! I blame the panic :P I have MATHEMATICAL as 12,0,19,7,4,12,0,19,8,2,0,11 –  Fred Mar 1 '13 at 12:14
Sorry, I don't do fast help. –  Gerry Myerson Mar 1 '13 at 12:18
I may be wrong, but the part "plaintext message units are length 2 and ciphertext units are length 3" seems to ask you to pack every two integers from the plaintext stream into a single RSA input $m \mapsto c = m^e \pmod{n}$. The RSA output $c$ is then break into 3 integers in $0, \ldots, 25$ and send to output in corresponding output alphabets. –  achille hui Mar 1 '13 at 13:03
@achillehui, since the modulus is 943, all ciphertexts will have 3 digits. –  mikeazo Mar 2 '13 at 2:05

RSA encryption works as follows, you have a modulus $n$ (in this case $943$) and an encryption exponent $e$ (in this case $3$). To encrypt a plaintext message $m$ you compute $c=m^e\bmod{n}$. So, if $m=12$, we get $12^3\equiv 785 \bmod{943}$.