Question

The ciphertext message produced by the RSA algorithm with \(\large (n,~k) = (1643,~223)\) is

\[\large \text{0833 0823 1130 0055 0329 1099} \]

Break the cipher first then decrypt the above message

Answer 1

lol can we cheat and have wolfram factor 1643

Answer 2

ofcourse yes! after all RSA is all about making things computationally infeasible..

Answer 3

n=p*q=31*53
totient(n)=30*52=1560
ed congrent 1 (mod 1560)
e is the public key
so I take your k to be my e
223d congruent 1 (mod 1560)
we can solve this for the private key d

Answer 4

\[223d-1=1560k \\ 223-1560k=1 \\ 1560=223(6)+222 \\ 223=222(1)+1 \\ 223-222=1 \\ 223-(1560-223(6))=1 \\ 7(223)-1560=1\]

Answer 5

\[d=7\]

Answer 6

\[m=c^d (\mod n) \\ m=c^7 (\mod 1643)\]

Answer 7

is the decryption function

Answer 8

Wow! I was thinking it would be tough but you're making it look too simple haha

Answer 9

decryption key = 7 is right !

Answer 10

0833 this is one member of the ciphertext right ?
\[m=833^7 (\mod 1643)\]

Answer 11

Yes and yes !

Answer 12

we may use wolfram to reduce that i guess

Answer 13

sorry my computer wasn't loading this thingy

Answer 14

yeah I don't know i non-wolfram way to compute that last thingy i said

Answer 15

like I would use repeated squares but 833^2 is ugly

Answer 16

using calculator is fine, yeah there is no fun in doing it manually

Answer 17

http://www.wolframalpha.com/input/?i=833%5E7++mod+1643

Answer 18

\[c=0833 \text{ } 0823 \text{ } 1130 \text{ } 0055 \text{ } 0329 \text{ } 1099 \\ m=170 \text{ } 415 \text{ } 112 \text{ } 499 \text{ } 131 \text{ } 422 \]

Answer 19

I got the same xD
i am using below site to convert numbers to alphabets
http://cryptoclub.math.uic.edu/substitutioncipher/ltr_num.htm

Answer 20

170 415 112 499 131 422

grouping 2 each and replacing 99 with space :

```
17 04 15 11 24 13 14 22
```

Answer 21

17=r
04=e
15=p
11=l
24=y
13=n
14=o
22=w

reply now

you made this up?

Answer 22

Nice :) Nope it is a textbook exercise.. Honestly I didn't expect it to be too easy lol ty :)

Answer 23

well I kinda looked at the RSA some back in college
and then also a calculator helps :p

Answer 24

i made something like this a long time ago
i will see if i can find it

Answer 25

the textbook encourages to use calculator for all exercises in cryptography
please id like to very much go over all interestig crypto problems xD

Answer 26

ciphertext:
15 02 25 17 25 , 01 17 25 , 52 03 , 11 04 09 49 24, 05 41 , 36 25 05 36 23 25 , 47 , 15 02 05 24 25 , 12 02 05 , 42 01 09 , 21 09 49 25 17 24 15 01 09 49 , 18 04 09 01 17 20 , 01 09 49 , 15 02 05 24 25 , 42 01 09 , 42 01 09 32 15 , 46



now "words" are separated by ,
words can be just a punctuation mark

m=01-26 is the normal A-Z
m= 27 you have 00
m=28 you have 01
m=38 you have :
m=41 yu have .
m=43 you have '

Answer 27

I'm trying to piece what the origial problem was based on pieces

Answer 28

lol I still haven't gave you public key yet

Answer 29

(n,e)=(55,27) is what is published to the public

Answer 30

yes im waiting haha
and that code is pretty common i think... my textbook uses the same code..

Answer 31

Now you are also suppose to find any type-o 's in the above sentence. Don't worry there isn't many.

Answer 32

But the sentence above is a computer science joke.

Answer 33

Let me try and find the decryption key first

n=p*q=5*11
totient(n)=4*10=40
ed congrent 1 (mod 55)
e is the public key
so I take your k to be my e
27d congruent 1 (mod 55)
we can solve this for the private key d

Answer 34

\(27d \equiv 1 \pmod{55} \implies d \equiv \dfrac{1}{27}\equiv \dfrac{2}{54}\equiv \dfrac{2}{-1}\equiv -2\equiv 53 \pmod{55}\)

so \(d = 53\)

Answer 35

oh wait

Answer 36

ed congruent 1 (mdd 40) right?

Answer 37

mod*

Answer 38

Oops yes my mistake~
\[27d \equiv 1 \pmod{40} \implies d \equiv \dfrac{1}{27}\equiv \dfrac{1}{-13} \equiv \dfrac{-3}{39}\equiv \dfrac{-3}{-1}\equiv 3\pmod{40} \]
so \(d = 3\) ?

Answer 39

c = (15 02 25 17 25 01 17 25 52 03 11 04 09 49 24 05 41 36 25 05 36 23 25 47 15 02 05 24 25 12 02 05 42 01 09 21 09 49 25 17 24 15 01 09 49 18 04 09 01 17 20 01 09 49 15 02 05 24 25 42 01 09 42 01 09 32 15 46)

m = (20 8 5 18 5 1 18 5 28 27 11 9 14 4 19 15 6 16 5 15 16 12 5 38 20 8 15 19 5 23 8 15 3 1 14 21 14 4 5 18 19 20 1 14 4 2 9 14 1 18 25 1 14 4 20 8 15 19 5 3 1 14 3 1 14 43 20 41)

Answer 40

THERE ARE 10 KINDS OF PEOPLE: THOSE WHO CAN UNDERSTAND BINARY AND THOSE CAN'T.

i heard this before nice one ;)
http://archive.oreilly.com/pub/post/there_are_10_kinds_of_people_e.html

Answer 41

lol