20100401, 10:18  #1 
Mar 2010
On front of my laptop
7·17 Posts 
My new discovery!!
Decrypt that RSA encrypted sentence to find out!
Encrypted sentence: 516706041301498569139808807078772714750649530964326734 N: 653465099972074208154022739130081051703785966122945327 E: 6725175846738058345386447704209495270349 10~35: A~Z 36: , 37: . 38: ? 39~48: 0~9 49: ! 99: (space) 
20100401, 10:50  #2 
Mar 2010
On front of my laptop
7·17 Posts 
Whoops!
Someone please move this thread to Lounge?

20100401, 11:02  #3 
Noodles
"Mr. Tuch"
Dec 2007
Chennai, India
2351_{8} Posts 
This is a good puzzle
102527182199152424214949 APRIL FOOL!! N = p.q p = 751567152453117379356088079 q = 869470010549506074668255713 phi = (p1)(q1) D = E^{1} (mod phi) such that E.D = 1 (mod phi) x = y^{D} (mod N) y = x^{E} (mod N) where x is the plain text y is the cipher text Good work man!! Where did you get up those all of your keys from, in order to generate that up only, thus? 
20100407, 11:46  #4  
Mar 2010
On front of my laptop
7×17 Posts 
Quote:
2. I entered it to Dario Alpern's ECM applet to find a prime. 3. And I used Dario Alpern's another applet to do the rest. 

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