Question

Discrete math 1

1) In RSA algorithm, suppose the public key n = 55 (11*5) and e = 17. Please find a private key d which ed = 1 (mod 40) (40 = (11-1)(5-1))

2) From the previous problem, suppose we have a message "one" (Letter A to Z are coded as 00 to 25), what will be the ciphered message?

3)Rewrite the classic Binary Search method as a Recursive Function (just the pseudocode.)

4)Deduce that, if A ⊆ B, B ⊆ C and C ⊆ A, then A = B = C.

Answer 1

A-60 Bto .: nth letter will be coded as (n-1) in 2. digit Code. "one" + "141304" Benaly search method. Procedure bsearch CA, Start, end, val) begin Hof ( Start L-end) mid;: (staut + end) 72 14 CA[mid) == val) return mid end it it (A (mid]> hal) return bsearch (A, Stalt, middle-l, val) end if it (A[id] cval) return bsearch CA, mid, end, val) end it endit return-l end BCC and CCA 4) A CB then A=B=c ACBC ECA 9 17 MEA -) MEB CACB] Also ne B = NEC [. Bec] SO, ME A =) HEC -> ACC ( As t is achttravily chosen] Acc in and cs Scano Ain camescanficet > Acc в ссс A 2 B C A AB; A=B=0

Given that the e-17 public key n=55 C11A5) 8.(n) = 40 GI 40 { it is not a perfect integery - Perfect integee d= 560+ 1 33 --) CCP) 134 mods we need to find ed = 1 mod 8(n) 17 xd i mod 40 d= 40+1 19 d- 801) 19 17 d=33 Matching with alphabets: A B CD E F G H I J K L M 23 4 5 6 7 8 9 10 12 13 RS T х Y Z 17 18 19 20 21 22 24 25 we need to find out cc = me moda ONE 19 n=13 e: 4 =) CC = 4 17 mod 55 49 mod as = 2464) ciphered о РФ 14 15 16 23 ciphered message v=14 ») CC = 14 mod 55 9(5) 18 (s) 59 message & lysy S CS Scani ed with CamScanner in