Public Key - RSA

Number Theory Level 2

Friends are exchanging messages via code. They decide to use ASCII alphabet where the letter $A=65$ and $Z=90$ . Everything else is in order and between these as usual.

There is a public key, $n=977770777$ and $e=13$ and both friends are aware of this. The sender converts their message into the ASCII alphabet so it becomes a number, let us denote this by $m$ .

The sender then takes the message, raises it the e, and looks upon it Modulo n. This is the ciphertext.

$c=m^{e}\equiv 150095694 \pmod{n}$

So the receiver gets the message $150095694$ and decrypts it back into ASCII and then back into the original text.

Decode the ciphertext into the original message and give the answer as the sum of the values of each letter in the regular English alphabet where A=1 and Z=26.

The answer is 56.

1 solution

William Allen
Mar 5, 2019

There exists a $d$ such that $ed \equiv 1 \pmod{\phi(n)}$ So as $m^{\phi(n)} \equiv 1 \pmod{n}$ we have $m^{k\phi(n)+1} \equiv m\times (m^{\phi(n)})^{k} \equiv m \pmod{n} \ \forall k\in \mathbb{Z}$ .

So we need $m^{ed} \equiv c^{d} \pmod{n}$

$\phi(n)=977705316$ so $d=300832405$ .

$c^{d}\equiv 87737676 \pmod{n}$ so putting this back through ASCII spells WILL.

So we have $23+9+12+12=\boxed{56}$

Public Key - RSA

The answer is 56.

1 solution

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