Question

I am implementing my own RSA algorithm in ruby to learn more about the algorithm.
It is almost working, but when I decrypt it a few numbers don’t decrypt but most do.
Why would this be?

For the given plaintext:
[ 42, 24, 424, 224, 421, 321]

The ciphertext is:
[1239,1263,1495,1349,208,1878]

Which when decrypted is:
[42,690,424,779,421,321]

This is the problem. Why is it happening?

These values are used to produce the keys (The method call is at the end of the program)

p = 51
q = 37
e = 223
n = 1887
phiN = 1800 (coprime with d)
d = 1687




class RSA
#edited to be concise, such as omitting initialize()

def encrypt(plainText)
index = 0
puts 'encrypt in is ', plainText
    plainText.each do |i|
    plainText[index] = ((i**(@e)) % @n )
    index+=1
    end
 puts  'ciphertext is '
puts plainText
return plainText
end


def decrypt(cipherText)
puts 'decrypt in is ', cipherText
index = 0
    cipherText.each do |i|
    cipherText[index] = ((i**(@d)) % @n )
    index+=1
    end
puts 'plaintext is '
puts cipherText
return cipherText
end


def calcD()
@d=1
begin 
s = (@d*@e)% @phiN;
@d+=1
end while not s==1
@d -= 1
#puts 'd is ', @d
end


end # class

message = RSA.new(51,37,223,[ 42, 24, 424, 224, 421, 321])

Answer 1

51 is not a prime.

Since that is one of the assumptions for the RSA algorithm, it should be no surprise that it fails to work.

Since your p is not a prime phi(n)!=(p-1)(q-1).

You could make it work by noticing that phi(51*37)=phi(3*17*37)=(3-1)(17-1)(37-1)=1152 and then calculating a working d=e^-1 (mod phi(n)) = 223^-1 (mod 1152) = 31, but I would recommend just using a prime p instead.