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• Dicska@lemmy.world ⁨1⁩ ⁨year⁩ ago

  Nah, they just used the entire number.

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• jaybone@lemmy.world ⁨1⁩ ⁨year⁩ ago

  That’s just an approximation though right? I doubt the number has that many zeros.

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  • scharf_2x40@lemmy.world ⁨1⁩ ⁨year⁩ ago

    Yeah it is, the exact number is 25104128675558732292929443748812027705165520269876079766872595193901106138220937419666018009000254169376172314360982328660708071123369979853445367910653872383599704355532740937678091491429440864316046925074510134847025546014098005907965541041195496105311886173373435145517193282760847755882291690213539123479186274701519396808504940722607033001246328398800550487427999876690416973437861078185344667966871511049653888130136836199010529180056125844549488648617682915826347564148990984138067809999604687488146734837340699359838791124995957584538873616661533093253551256845056046388738129702951381151861413688922986510005440943943014699244112555755279140760492764253740250410391056421979003289600000000000000000000000000000000000000000000000000000000000000000000000000000000000000000

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    • jaybone@lemmy.world ⁨1⁩ ⁨year⁩ ago

      Thanks! I’m still surprised at the number of trailing zeros though. Is that a function of how many multiples of ten there are in the factorial?

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      • Natanael@slrpnk.net ⁨1⁩ ⁨year⁩ ago

        Just looked it up

        www.purplemath.com/modules/factzero.htm

        If we take the factorial of any number larger than 5, then there will be at least one zero at the end of the number. Why? Because 5! = 1×2×3×4×5; in particular, 5! = (2×5)×(1×3×4), and (2×5) = 10. The factorial of any larger number will have more copies of 2 and 5 (as factors of larger values, like 6 and 15), so there will be even more factors of 10 in these factorials. And every factor of 10 adds a zero to the end of the factorial expansion.

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