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Absolutely certain!Are you sure You don't mean 0x16?
I don't get it.Absolutely certain!
Err, wait ... I typo'd the number -- it should be 0z15, which == 0x29
105 97 109 51 57
Figure that one out
Work it out ... 41(b10) = 15(bX) ... simple mathI don't get it.
So you're 41, but what base is 0z?
It's not base 32 because 1s are not used there...
Except that notation in base 32 is different from what you might expect.Cy said:If the base was 32, the number would be 32^1*(second digit) + 32^0*(first digit) = 41
which is 0z19.