madeeasy test series

Question

Two gate aspirants talking to each other use the RSA algorithm to encrypt their messages. They encrypt the message character by character. The value of p, q and d are 5, 17 and 13 respectively, where p, q and d are their integers having usual meaning in the RSA algorithm. Assume that corresponding cipher characters are placed in their corresponding plain text character places. Also each character is converted to ASCII value before applying RSA (ASCII value of A, B, C,... and so on are 1, 2, 3,..., respectively).The sum of integers in cipher text for corresponding characters in plain text: “IIT” _________ .

Comments

I am getting 198(I=59,I=59,T=80)

Answer 1

SUMMARY OF RSA :

1. Choose two very large prime numbers p and q.
2. Multiply them n = p X q.
3. Calculate Θ=(p -1) X (q - 1).
4. Choose a random integer e. Then calculates d so that d x e= 1 mod Θ
5.  e and n are public;  Θ and d are private.

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In given problem p= 5, q=17 then n= 85

Θ =(5-1)*(17-1)= 4*16 =64

d x e= 1 mod Θ means 13 x e = 1 mod 64

Since 64*1+1 =65 is a multiple of 13 (Since 13 x 5 =65) , e=5

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Sender encrypts as: C=P^e (mod n)

Given plain text is IIT, ciphertext is

C( I) = C (9) = 9 ^5 (mod 85) = 59

C(T)= C(20) = 20 ^5 (mod 85) =5

Sum of integers = 59+59+5 =123

Comments

I think e should be 5. Right?

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Yes.. Corrected.!!!

Thank u for pointing out ..:)

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yes e will be 5

so 9^5 mod 85 = 59

and 20^5 mod 85 = 5

so answer will be 59+59+5 =123 is answer

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public key is used for encryption?? then how come receiver will decrypt it when secret key is also with sender only?

i mean we should have encrpted I with

C(I) = C(9) = 9^13 mod 85

????

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If sh!va uses RSA he announces e and n to the public; but keeps Θ and d secret.

Anybody can send message to sh!va with encryption as: C=P^e (mod n)

Sh!va  decrypts it  as: P= C^d(mod Θ)

I hope concept is clear now

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yes.. :)

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Is there any trick to solve remainders including higher power?

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inverse multiplication

Answer 2

I got 123...

Answer 3

123 is the answer. e=5 as 5*13 mod 64=1  for I = 9^5mod 85=59, for T= 20^5mod 85= 1, sum = 123 (59+59+1)

Comments

20^5 mod 85 = 1, then why we are writing it as 5 .. please reply

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$20^{5}=3200000 \text{mod} 85=5$