Question

Use the RSA cipher with public key n = 713 = 23 · 31 and e = 43.
(a) Encode 3 and 15 to find their cipher text.
(b) Find the least positive inverse for 43 modulo 660 which is the decryption key d.
(c) Decode the plaintext for 28, 18 and 129.

Answer 1

Given n=713, p=23, q=31 and e=43

1 < e < (p-1)(q-1) and n = p * q

1< 43 < 660 and 713 = 23 * 31 (hence all values are correct)

Public Key =>(713,43) as (n,e)

Now, Generate the private key,

we know the formula

to find the value of d, we will use the Extended Euclidean Algorithm

step-1) Euclidean Algorithm

((p-1)*(q-1))=e*x+r 1) 660=15(43)+15 2) 43=2(15)+13 3) 15=1(13)+2 4) 13=6(2)+1

step-2) Back Substitution

1=13-6(2) ( from 4) 1=13-6(15-1(13) (from 3) 1=13-6(15)+6(13)=7(13)-6(15) 1=7(43-2(15))-6(15)=7(43)-14(15)-6(15) (from 2) 1=7(43)-20(15) 1=7(43)-20(660-15(43))=7(43)-20(660)+300(15) (from 1) 1=307(43)-20*660

now, we know

hence d=307

Private Key-> (307)

(Formula for Cipher conversion)

3 will be encoded as 675

5 will be encoded as 89

Here is the least positive inverse for 43

Euclidean Algorithm ----------------------------- ((p-1)*(q-1))=e*x+r 1) 660=15(43)+15 2) 43=2(15)+13 3) 15=1(13)+2 4) 13=6(2)+1

Back Substitution ------------------------------------- 1=13-6(2) ( from 4) 1=13-6(15-1(13) (from 3) 1=13-6(15)+6(13)=7(13)-6(15) 1=7(43-2(15))-6(15)=7(43)-14(15)-6(15) (from 2) 1=7(43)-20(15) 1=7(43)-20(660-15(43))=7(43)-20(660)+300(15) (from 1) 1=307(43)-20*660

Hence, least positive inverse for 43 is 307

for 28 -------- x= Ans is 14

for 18 -------------- x= Ans is 9

for 129 ------------------ x= Ans is 5

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