Saturday, 1 February 2014

elementary number theory - Decrypting an Affine Cipher e(m)=am+bpmod27 knowing e(8)equiv14 and e(26)equiv5



I began by setting up a system of linear equations:
148a+b(mod27)
526a+b(mod27)
and then subtracted them to get: 99a(mod27). I know 9 doesn't have a multiplicative inverse modulo 27 but a=1 would solve this. I'm afraid I must be making some mistake somewhere, however, as solving this all the way through gives me b=6 and then translating the message doesn't quite make sense.


Answer



9a9mod27
a1mod3

a=1+3k
b148amod27
b=148(1+3k)+27m
b=624k+27m
Where k,mZ.


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