Step 1: F < 10,000,000. Proof: Any factorian with N digits cannot exceed 9! * N. Since 9! = 362,880, 9! * N has less than N digits for all N > 7. So, any factorian must have at most 7 digits.
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Step 2: F < 2,000,000. Proof: Suppose F > 2,000,000. Since F has 7 digits, F is at most 9! * 7 = 2,177,282. So, F’s 1st digit must be “2” and 2nd digit must be “1” or “0”. But then its factorial sum cannot exceed 2! + 1 + 9! * 5 = 1,814,403, less than 2,000,000, a contradiction!
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P.S. Interested readers can build on this argument to show that F < 1,500,000. Hint: F’s last digit cannot be “9” unless it contains a “1” or “0” and at least two more digits less than “5”.
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Whenever I see questions dependent on base 10, I wonder how to generalize the question to make it base-agnostic.
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Factorions in other bases: http://oeis.org/A193163
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Usefulness shows up in surprising ways, usually at surprising times, and mostly not in one's lifetime. And yet it does appear. An application of the properties of large numbers.
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having fun with math is useful. other than that, unlikely :)
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Factorion not factorian?
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you don't want to stress yourself if doesn't have useful application ?Thanks. Twitter will use this to make your timeline better. UndoUndo
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