Solution to Encrypt the message ATTACK using the RSA cryptosystem with n = 43 · 59 and … - Sikademy
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Archangel Macsika

Encrypt the message ATTACK using the RSA cryptosystem with n = 43 · 59 and e = 13, translating each letter into integers and grouping together pairs of integers, as done in example 11 in the textbook and in the classnotes

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RSA uses a public key, found from the product of 2 large primes. In yhe given case it is they selected 43 and 59, so the public key is 43*59 = 2537. You were also given e = 13. The English alphabet contains 26 letters, and there is two ways to mark the letters: from 0 to 25 or from 1 to 26. Lets mark them from 1 to 26(A - 1, B - 2 etc). The logic of encrypt next: We transform each letter into the numeric form using alphabet above(A - 1, T - 20, C - 3, K - 16) and encrypt it using the next formula: m=k^emod(q) , where m - encrypted letter, m - numerical form of the letter, e - public exponent, q - public key

In the given case:

A = 1^{13}(mod2537) = 1

T = 20^{13}(mod2537)

20^{3}(mod2537)=389\to 20^{4}(mod2537)=20*389(mod2537)=169\to\ ...\to 20^{13}(mod2537)=1435

С=3^{13}(mod2537)=1087

K=16^{13}(mod2537)=1225

So, the encrypted message is 114351435110871225

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