3/7     How do I solve for a and b (they are strings)?
        f(a) = 2.90171130144904e+22
        f(b) = 24,386,094,146,700
        where
        f(x) = y=1;for(i=0; i<x.length; i++) {y *= x.charCodeAt(i);}
        \_ If you want to find nice-looking strings, you could do a
           dictionary search -- try encoding lots of different plausible
           strings until you find some that give you the values you want.
           I like "paidsurveys" and "user567".  --mconst
        \_ if you were more than a half-wit, you would have realized that
           there are infinitely many solutions, or at least a lot, depending
           on what alphabet your strings are over
        \_ As per above, there _are_ a lot of solutions (trivially, a finite
           number since the above in all likelihood is over ASCII). However
           that doesn't mean that this is _completely_ impossible. You _do_
           however need to explain more about the problem. What do you know
           about a and b? What kind of characters can they possibly contain?
           Upper-case letters? Lower-case letters? Punctuation? Digits?
           Control characters? "Foreign"/"special" characters (a la extended
           ASCII, 0x80 and up)? What are the strings supposed to be? Words?
           Do they follow some interesting statistical distribution? ...Or,
           alternatively, do you need a _specific_ solution out of many,
           or will any string that f returns those values on suffice?
           Also, you make your job _far_ harder by giving floating point
           forms for the first one; since you're missing several digits from
           the end of the decimal representation, you'll effectively have
           trying to get into?  -alexf
           to make up for it by considering all possible values for them
           (making your job literally 10,000,000 times harder under most
           circumstances). Also, since the format of your statement suggests
           that this is javascript, which shittily-secured porn site are you
           trying to get into? We want some too =)  -alexf
           \_ P.S. if said strings don't contain extended ascii chars, I can
              at least tell you that b contains exactly one "e", and no "a"'s,
              "g"'s, "k"'s, "m"'s, or "q"'s. See also: prime factorization =)
              \_ of course, it should go without saying that mconst is
                 infinitely smarter than i, and you should just go read his
                 post above instead. -alexf