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?

@David Holcer
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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 \(k\).

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## Comments

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TopNewestI crashed my computer doing this ... Nice question by the way

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No output in 4 hours!! :(

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Great question to ask!

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Well I tried the same thing with S(n)>S(n^4) and it didn't work either.

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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

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Oh It seems my computer can't handle such large numbers.

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