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Alice and Bob have agreed to use the Three Pass Protocol.

p=1009

Alice chooses the encryption exponent

e_A = 101

Bob chooses the encryption exponent

e_B = 209

Now Alice and Bob send three encrypted messages: y1, y2, and y3 in this order.

Which of the following are possible solutions?

a. y1 = 471, y2 = 27, y3 = 14 is possible

b. y1 = 27, y2 = 471, y3 = 14 is possible

c. y1 = 471, y2 = 14, y3 = 27 is possible

d. y1 = 27, y2 = 14, y3 = 471 is possible

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  • $\begingroup$ Show us what approaches have you tried. $\endgroup$ – mentallurg Feb 22 at 0:20
  • $\begingroup$ @mentallurg I've been trying to use the modular inverse to solve it in mathematica using trial and error. For example. PowerMod[471, 209, 1009] = 14 and then PowerMod[14,101,1009] = 548. That is incorrect and I'm not sure if even my method is correct. $\endgroup$ – sherrie1618 Feb 22 at 0:31

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