Points$]$

EPPDISCMATH$58\mathrm{.}4\mathrm{.}024.$

Alice received the following ciphertext from Bob, $"081408".$ Bob had encrypted it using the RSA cypher with Alice's public key $(pq,e)=(55,3),$ where $p=5$ and $q=11.$ Note that $(p-1)(q-1)=40.$ The value for $d$ in Alice's private key, $(pq,d)$ is a positive inverse for $3$ modulo $(p-1)(q-1).$ It was found to be $27$ in Example $8\mathrm{.}4\mathrm{.}8($b$)$ and Example $8\mathrm{.}4\mathrm{.}10.$ Wha message after Alice decrypts it$?($Assume Bob encoded one letter at a time using the encoding $A=01,B=02,C=03,$dots,$Z=26.)$

To decrypt Bob's message, Alice uses the decryption formula

$M=C,$mod

where $M$ is the code for a letter of the message, $C$ is the encrypted version of the letter, $(pq,e)=(55,3)$ is the public key, and $(pq,d)=(55,27)$ is the private key.

$($a$)$ To begin, Alice computes the values of $a,b,c,d$ and $e$ that are indicated below.

${08}^1-=a(mod55),{08}^2-=b(mod55),{08}^4-=c(mod55)$

${08}^8-=d(mod55),{08}^{16}-=e(mod55)$

She finds that $a=b=c=,d=,$ and $e=$

Because

$27=16+8+2+1,{08}^{27}={08}^{16+8+2+1}={08}^{16}*{08}^8*{08}^2*{08}^1,$

she uses the values of $a,b,d,$ and $e$ to compute ${08}^{27}$mod$55=(a*b*d*e)mod55=$

Thus, the first letter in Bob's message is $$

$($b$)$ Alice finds the second letter of Bob's message by computing $27$mod$55=$

$($c$)$ What is Bob's message after Alice finishes decrypting it$?$

$$