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.

${51}^1-=a(mod55),{51}^2-=b(mod55),{51}^4-=c(mod55)$

${51}^8-=d(mod55),{51}^{16}-=e(mod55)$

She finds that $a=1$

$,b=$

Because

$27=16+8+2+1,{51}^{27}={51}^{16+8+2+1}={51}^{16}*{51}^8*{51}^2*{51}^1,$

she uses the values of $a,b,d,$ and $e$ to compute ${51}^{27}$mod$55=(a*b*d*e)mod55=$

Thus, the first letter in Bob's message is $23$

$($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$?$