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Nanashi comments on A pair of free information security tools I wrote - LessWrong
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Comments (97)
Why does it need to be a multiple of 3?
(SHA-2 = 2^256 = 1*10^77)
You wrote that the odds were 1 in 1 quattuorvigintillion. I was under the impression that all "-illion"s have exponents that are multiples of 3.
Ahhhh. I misread the output on Wolfram Alpha. You're right. I'll leave it in the original post for posterity, but also for the record, it's actually 1 in 100 quattuorvigintillion
(That's what I get for trying to be dramatic)