TITLE: Public Key Cryptography
QUESTION [5 upvotes]: Assuming that a message has been sent via the RSA scheme with $p=37$, $q=73$, and $e=5$, what is the decoding of the received message "34?"
So far, I have $x^5 \pmod{37\times73} \equiv 34$.  How do I invert a mod?

REPLY [2 votes]: We need the totient function of the modulus, hence we get:
$$\varphi(N) = \varphi(37 \times 73) = 36 \times 72 = 2592$$
First you need to find the decryption exponent.
$$d = e^{-1} \pmod {\varphi(n)} = 5^{-1} \pmod {2592} = 1037$$
Decryption:
$$\displaystyle m = c^{\large d} \pmod N \rightarrow 34^{1037} \pmod {2701} = 1415$$