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Great. I would like to see the mathematical proof of this assertion. Any takers?
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Just prove there is a finite number of them then brute force
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It only works in base 10 though...
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There are always 10 responses to all Fermat's Library fun facts. The first is "Proof please!" and the second is "Rubbish as it only holds for base ten." Even now some of you will want proof of my assertion, and others will say..."Ah, never mind. He wrote this in base two!"
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Watch how our handwriting
#Calculator operates and does this calculation:pic.twitter.com/BRf0hNv5mDThanks. Twitter will use this to make your timeline better. UndoUndo
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Explain me these, 0000000010 000000011 00000000012 00000013
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OK, let's just assume this is cheating x)
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You know, that's a bit like the 2 first Fibonacci numbers being the same, except it's not the same reason and therefore my explanation makes no fucking sense.
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Lemma for the proof outlines described... Let N have M digits with M >= 100. Then N! is greater than M x 9!. (Prove by induction with M_0 =100.) This would establish that the solution space is finite and bounded by essentially a googol. Overkill to be sure, but it works!
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A googol? Sheeiiyt. I tackled the proof like this (in base 10): Take a set of M, where M_1=9, M_2=99, M_x=...). Set N is output (sum of M_x respective digit factorials). Goal is to find an x for which: length(M_x) > length(N_x) and you will find a solution space. I found x=8...
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