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No S(n)>S(n^5)?

Inspired by this and this, I tried the same thing with n^5 and couldn't find an x that worked (or it is much much larger). Is there an x such that the sum of the digits of x is greater than the sum of the digits of x^5?

Note by David Holcer
2 years ago

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I crashed my computer doing this ... Nice question by the way Soutrik Bandyopadhyay · 2 years ago

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No output in 4 hours!! :( Pranjal Jain · 2 years ago

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Great question to ask! Calvin Lin Staff · 2 years ago

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Well I tried the same thing with S(n)>S(n^4) and it didn't work either. Ronak Agarwal · 2 years ago

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@Ronak Agarwal 124499 works for the forth power. 1+2+4+4+9+9=29 124499**4=240250031031001002001. 2+4+2+5+3+1+3+1+1+2+1=25 David Holcer · 2 years ago

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@David Holcer Oh It seems my computer can't handle such large numbers. Ronak Agarwal · 2 years ago

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@Ronak Agarwal Depends on programming language you are working upon Pranjal Jain · 2 years ago

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@Pranjal Jain Do you think there is a way to prove whether or not there exists an x for x^5? David Holcer · 2 years ago

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@David Holcer Probably there exists some large \(x\) for which the inequality is satisfied. Probably finite solutions exist only when you consider \(x^{k}\) for very very large \(k\). Jake Lai · 2 years ago

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@Jake Lai Clearly if there is one such \(x\) then there are infinitely many such \(x\); just append zeroes at the end. Ivan Koswara · 2 years ago

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@Ivan Koswara Yes and this proves that either there are no solutions or there are infinite solutions. Ronak Agarwal · 2 years ago

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@Ivan Koswara Yeah, but I mean one is not a power of ten times of the other. Otherwise, it would be pretty boring, wouldn't it? Jake Lai · 2 years ago

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@David Holcer Not till now Pranjal Jain · 2 years ago

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@Pranjal Jain What, what number?!! David Holcer · 2 years ago

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@David Holcer I mean I don't know if there is any way to prove whether such n exists or not Pranjal Jain · 2 years ago

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@Pranjal Jain Oh, ok.Thnx. David Holcer · 2 years ago

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