et n= 23 t ng 13 Share En, nz =299and e=133.if e message is m=3 then encrypt Given messase via the RSA çacithm to the enerypted meSsa ge n) =me mod 299 Then find. decryptien ekRenent d and show (m) y =m Em mod 299.

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A: There is no question in this, Please repost the question with which algorithm we need to decode it.

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Q: et ni = 23 and n2 = 13. Share n = n¡n2 = 299 and e = 135. If the message is 2 = 3 then encrypt the…

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Q: Use the encryption congruence C = (3P + 7) mode 26 to encode the message EASY

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Q: = 1010001101. The CRC f e using the divisor polynor

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