"Crypto paper Section C proofing + Update on life"

First of all, we will be discussing on Asymmetric , Public Key Encryption : RSA.

So, how RSA works? :

We will need two (very) long prime numbers ; p and q

For the sake of simplicity, we will choose 3 and 11 as our prime number. Also, these numbers must be kept secret.

Now, we will calculate n = p * q

n = 3 * 11 = 33

n value should be made public.

Now, we will calculate what we call the totient funct. from Euler : φ(n) = (p-1)(q-1)

φ(n) = (3 - 1)(11 - 1) = (2)(10) = 20

Next, we can derive the public key from this equation : e, with gcd(φ(n), e) = 1 ; 1 < e < φ(n)

e value should be made public since it's public key. This key is used to encrypt messages, on sender side.

Let's say hypothetically e = 7

Next, we can derive the private key from this equation : d = e^-1 (mod φ(n))

If value of e or d is already derived, then this equation should be used : de mod φ(n) = 1. Using simple algebra, we need to find values of e or d that conform to this equation.

de mod φ(n) = 1 ; d(7) mod 20 = 1 ; (3)(7) mod 20 = 1 ; d = 3

d value must be kept private as it is the key to decrypt messages that was encrypted by the public key, on receiver side.

Now... lets see the question on my final paper, Semester 2 2022/2023.

p = 61 ; q = ? ; n = 3233 ; e = 17 ; d = ?

So we plug in our previous equations (with simple algebra):

n = p * q ; 3233 = 61 * q ; q = 3233 / 61 = 53

φ(n) = (p - 1)(q - 1) = (61-1)(53-1) = (60)(52) = 3120

e has already derived = 17

de mod φ(n) = 1 ; d(17) mod 3120 = 1 ; (2753)(17) mod 3120 = 1 ; d = 2753

Now, here where is the problem with large numbers calc. By the rule, in order to encrypt message : Ct = M^e mod n and to decrypt message : M = Ct^d mod n

Following the rule, given a Ct = 2790, the equation should be : M = 2790^2753 mod 3120

That will yield : 5.6023677355623908532752104320819e+9485 mod 3120. That is obviously a comically large number for a Casio fx-570MS calculator to calculate. During the exam, it returned MATH ERROR.

However, continuing to mod the number, we get seemingly small value : 65

In symm. dec. , we are taught to use alphabetic-numerical table to derive the plaintext (0=A, 1=B, 2=C,...), but the value derived from the decryption process is not a garbage value either.

Instead, it corresponds to " A " ; using ASCII value and convert it to text.

To conclude, I still don't understand, why do large numbers are used for the derivation of the private key. There is probably a mishap in the making of the question itself. It's not thought through thoroughly. The calculation of power on large numbers is simply not possible on simple calculators.

However, I do appreciate that I (might) be the first to crack the Ct, if not.

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"Crypto paper Section C proofing + Update on life"

Now... lets see the question on my final paper, Semester 2 2022/2023.