I'm stuck with this problem: Assume Alice and Bob are using RSA to communicate each other. Given Bob's public key $(217,7)$, Alice send the ciphertext $c=53$ to Bob through RSA encryption. Now, Eve can read the ciphertext $c$, how can Eve decrypt the ciphertext and read plaintext $m$?

  • $\begingroup$ @user30025 Is this (let’s just call it) “homework”? What research have you done? What have you tried? $\endgroup$
    – e-sushi
    Dec 22, 2015 at 21:07

1 Answer 1


Some examples of answers to similar questions:

Hint: $N = 217 =\ ?\ \times\ ?$

In the public key, $e$ is the exponent and $N$ is modulus, therefore $e < N$.
If $m$ is your message, $m^e \mod N = c$ is the cipher.
To decipher you need to find $d$, such as: $d \times e = 1 \mod \phi(N)$.
You will then be able to find $m$: $c^d \mod N = m$

  • $\begingroup$ N = 217, public key = 7. You can find them, you know that p and q are prime. $\endgroup$
    – Biv
    Dec 22, 2015 at 20:55
  • $\begingroup$ public key is represented as (217, 7), can you please explain me what values are e and n? $\endgroup$
    – user30025
    Dec 22, 2015 at 21:00
  • $\begingroup$ 1 is not prime. it is N = p * q. Any method to find them is fine, you can even ask google if 217 is prime. :) $\endgroup$
    – Biv
    Dec 22, 2015 at 21:56
  • $\begingroup$ You need to factor $N$ not $e$. 7 is prime so it can not be a product of two primes. 217 ist not prime, but consists of two factors (excluding 1 and 217 itself). $\endgroup$
    – Kevin__
    Dec 22, 2015 at 21:58
  • $\begingroup$ If factoring $N$ was forbidden by the rule of the homework, of If $N$ was too large to factor, there is the option to try every possible plaintext, encipher it using the public key, and check against the ciphertext. $\endgroup$
    – fgrieu
    Dec 28, 2015 at 10:37

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