The reading for today was about the Chinese Remainder theorem which can be used to break a congruence problem into a system of smaller congruences, or to take a system of congruences to a single congruence statement. I assume this is how RSA works, by large number systems of congruence. But I need to reread that section to get it clear. I reread the section on block sipher modes and it makes sense now. I think that the reading-lecture-rereading is the best method of study.
But back to the CRT, if x = 4 mod 7 and x = 3 mod fifteen, then x=18.
Larger numbers are only more difficult because the process is longer, but not too hard.
Still kind of scared about the test though. Esspecially because I have class all day Tuesdays and will have to take the test tomorrow.
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