MATH 243 -- Algebraic Structures 

RSA Encryption/Decryption Example 

11/5/2012 

For this "toy example" we will use an RSA system with  

so the modulus is  As  

encryption exponent we need some  with  

We use  which is OK since   

Say the plain text message is  

     SEND TWO COPIES 

According to the basic encoding we have used this becomes 

      19,5,14,4,0,20,23,15,0,3,15,9,5,19 

For this RSA system, since the modulus is about 1000, we can 

group the numerical form of the plain text into three-digit  

"blocks" and apply the encryption function   

to each block like this: 

>

 

(1)

 

>

 

>

 

(2)

 

(This would not be converted back to a literal format for the RSA system -- 

the cyphertext is a string of numerical data.) 

Now, knowing we can compute the decryption exponent easily -- it is  

the    such that  in  By our usual methods, we know 

that is   

For a "real" RSA system, though, the security would come from the fact that  

the factorization m = pq  would not be known, so, even knowing the encryption  

exponent  e,  there would not be any obvious way to find in   

Here, we do have so we can decrypt easily (and parallel to the method 

used for encryption): 

>

 

>

 

(3)

 

When this is broken back up into two-digit groups and converted back to literal 

form, the original message is recovered(!)