rsa calculations

Question

I am trying to solve RSA for 2 small numbers. I can calculate n,phi and e but I always get stuck when I have to calculate d. Please help me with the same. example.

        p = 3,      q = 7,
        n =  3*7 = 21,
        phi(21)= 2*6 = 12, 
        e = 5

        d = (5^-1) (mod 21) 

        or

        d * 5 = k * 12 + 1   (where k is some number)

I tried to figure out the calculation of d * 5 = 25 = 5 * 12 + 1 but this is for small number is there any other way to calculate d with simple approach

what language are you using? What are you trying to do in your calculation of `d`? `(5^-1)` could be bit manipulation, to-the-power-of-'-1', ... — Floris, Mar 12 '13 at 15:00
"Please show some detail example" is what makes this question highly unsuited for Stack Overflow. — , Mar 12 '13 at 15:00
iirc, `(5^-1) (mod 21)` in this context means the inverse of 5 mod 21. — Reinstate Monica -- notmaynard, Mar 12 '13 at 15:04
Just started using Stack Overflow.. I beg ur pardon if I made some mistake — Kunj, Mar 12 '13 at 15:05
The reason your question is unsuitable here is because you've done the (very) easy calculations yourself and asked for us to do the difficult one for you, without showing that you've tried to figure it out yourself, or even that you've looked up how to do it. — Reinstate Monica -- notmaynard, Mar 12 '13 at 15:07
If you show that you've put forth real effort, we are much more inclined to help you. — Reinstate Monica -- notmaynard, Mar 12 '13 at 15:07
Sure. You might even see what went wrong and get the answer yourself. — Reinstate Monica -- notmaynard, Mar 12 '13 at 15:14
You could look at [this earlier question](http://stackoverflow.com/questions/3209665/for-rsa-how-do-i-calculate-the-secret-exponent) — Floris, Mar 12 '13 at 15:23

score 3 · Accepted Answer · edited Dec 26 '20 at 12:08

3

The following pseudo code may help you (with heavy borrowing from this link

// choose prime factors:
p = 3;
q = 7;

n = p * q; // =21
phi = (p-1)*(q-1); // = 12
// Choose e such that 1 < e < phi and e and n are coprime:
e = 5; 
// Compute a value for d such that (d * e) % phi = 1. 
// in other words, solve 5 * d % 12 = 1
d = 5; // since 5 * 5 = 25; modulo 12 = 1. How odd: d == e...

Public key is (e, n) => (5, 21) 
Private key is (d, n) => (5, 21) 

Testing this out on a "message" with the value of '2':
The encryption of m = 2 is 
    c = 2^5 % 21 = 32 % 21 = 11 
The decryption of c = 11 is 
    m = 11^5 % 21 = 161051 % 21 = 2

As you can see, we got the "message" back after the encryption / decryption step.

Note that since e==d, this is (unfortunately) a symmetric cypher: if you apply the encryption again, you get back the message as well. This shows that the choice of e was poor. That's a problem with these toy RSA problems...

edited Dec 26 '20 at 12:08

Toby Speight

27,591
48
66
103

answered Mar 12 '13 at 15:16

Floris

45,857
6
70
122

ok thanks for the help... also plz check the following – Kunj Mar 12 '13 at 15:32
@Kunj - "plz check the following"??? What do you mean? – Floris Mar 12 '13 at 15:40
1

ok thanks for the help... also plz check the following m = c^d mod 77 m = 10^37 mod 77 m = (67*67*67*7) mod 77 m = 7 ??? is this correct ? – Kunj Mar 12 '13 at 15:41
1

You might find [this calculator](https://www.cs.drexel.edu/~jpopyack/IntroCS/HW/RSAWorksheet.html) instructive. How did `10^37 mod 77` turn into `(67*67*67*7)mod 77`? You can use Wolfram Alpha's `PowerMod` [function](http://www.wolframalpha.com/input/?i=PowerMod%5B10%2C37%2C77%5D) -which gives `10` as the answer. – Floris Mar 12 '13 at 16:00

rsa calculations

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