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
–
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 \(k\).

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewestNo output in 4 hours!! :(

Log in to reply

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

Log in to reply

Great question to ask!

Log in to reply

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

Log in to reply

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

Log in to reply

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

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply