# How do I solve this RSA instance for m?

How we can solve this equation and get the value of M?

$$8 = M^{13} \mod 33$$

not a computer program, but a mathematical operation.

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## migrated from stackoverflow.comDec 13 '11 at 14:26

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Welcome to crypto! Your question was migrated here as it is more on topic here than on SO. Plus we have TeX! If you want to register your question will be associated with your account and you can receive notifications about it and reputation from it as normal, and be able to comment on it. – user46 Dec 13 '11 at 14:42

Factor the modulus, $n$ which was given as 33, to yield $p = 3$ and $q = 11$.

The totient, $\phi$ is $(p - 1)(q - 1) = 20$.

The public exponent, $e$ is given as $13$.

Now compute the private exponent $d$ as the multiplicative inverse of $e \mod \phi$, so $e^{-1}\mod \phi \equiv 17$.

The cipher text, $c$ is given as 8.

Finally compute the message, $M$ as $c^d \mod n \equiv 2$.

No algorithm has been published for factoring large numbers quickly, so the first step is what makes breaking RSA "computationally infeasible." Fast algorithms are known for all of the other operations.

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Since we're working modulo 33, we only need to look at values of M between 0 and 32. (Since $a^k \mod b \equiv (a \mod b)^k \mod b)$.

If we examine these 33 values, we get that it holds for M = 2.

However, this in fact holds for any $M = 2 + 33k$, where k is an integer, as per the argument outlined in the beginning.

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