# 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
5 years, 11 months ago

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No output in 4 hours!! :(

- 5 years, 11 months ago

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

- 5 years, 11 months ago

Staff - 5 years, 11 months ago

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

- 5 years, 11 months ago

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

- 5 years, 11 months ago

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

- 5 years, 11 months ago

Depends on programming language you are working upon

- 5 years, 11 months ago

Do you think there is a way to prove whether or not there exists an x for x^5?

- 5 years, 11 months ago

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 $k$.

- 5 years, 11 months ago

Clearly if there is one such $x$ then there are infinitely many such $x$; just append zeroes at the end.

- 5 years, 11 months ago

Yes and this proves that either there are no solutions or there are infinite solutions.

- 5 years, 11 months ago

Yeah, but I mean one is not a power of ten times of the other. Otherwise, it would be pretty boring, wouldn't it?

- 5 years, 11 months ago

Not till now

- 5 years, 11 months ago

What, what number?!!

- 5 years, 11 months ago

I mean I don't know if there is any way to prove whether such n exists or not

- 5 years, 11 months ago

Oh, ok.Thnx.

- 5 years, 11 months ago

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