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h0pe
 one year ago
Find an integer x such that 0 <= x < 527 and x^37 ≡3 (mod 527).
h0pe
 one year ago
Find an integer x such that 0 <= x < 527 and x^37 ≡3 (mod 527).

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UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0so we need a number x such that the remainder will be 3 in mod 527. Trial and error will take too long because the exponent is just too big here

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2start by factoring 527

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2that means \(\phi(527) = (171)(311) = 480\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2Now solve below equation \[37y \equiv 1 \pmod{480}\]

h0pe
 one year ago
Best ResponseYou've already chosen the best response.1I have to find the inverse of 37, right?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2that is, find the inverse of exponent "37" in mod 480

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2\[\large x \equiv 3^{13} \pmod{527}\]

h0pe
 one year ago
Best ResponseYou've already chosen the best response.1\[x\equiv1594323(\mod527)\] which is \[x\equiv148 (\mod527)\]?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2since they want x to be between 0 and 527, just pick x = 148

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2as you can see breaking RSA is trivial when you're able to factor the modulus

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2RSA is the name of an encryption system, I thought you're working on encryption algorithms... I went through your past questions...

h0pe
 one year ago
Best ResponseYou've already chosen the best response.1Oh, so that's what it's called..
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