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?

×

Problem Loading...

Note Loading...

Set Loading...

## Comments

Sort by:

TopNewestI crashed my computer doing this ... Nice question by the way – Soutrik Bandyopadhyay · 1 year, 10 months ago

Log in to reply

No output in 4 hours!! :( – Pranjal Jain · 1 year, 10 months ago

Log in to reply

Great question to ask! – Calvin Lin Staff · 1 year, 10 months ago

Log in to reply

Well I tried the same thing with S(n)>S(n^4) and it didn't work either. – Ronak Agarwal · 1 year, 10 months ago

Log in to reply

– David Holcer · 1 year, 10 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=25Log in to reply

– Ronak Agarwal · 1 year, 10 months ago

Oh It seems my computer can't handle such large numbers.Log in to reply

– Pranjal Jain · 1 year, 10 months ago

Depends on programming language you are working uponLog in to reply

– David Holcer · 1 year, 10 months ago

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

– Jake Lai · 1 year, 10 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 \(k\).Log in to reply

– Ivan Koswara · 1 year, 10 months ago

Clearly if there is one such \(x\) then there are infinitely many such \(x\); just append zeroes at the end.Log in to reply

– Ronak Agarwal · 1 year, 10 months ago

Yes and this proves that either there are no solutions or there are infinite solutions.Log in to reply

– Jake Lai · 1 year, 10 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?Log in to reply

– Pranjal Jain · 1 year, 10 months ago

Not till nowLog in to reply

– David Holcer · 1 year, 10 months ago

What, what number?!!Log in to reply

– Pranjal Jain · 1 year, 10 months ago

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

– David Holcer · 1 year, 10 months ago

Oh, ok.Thnx.Log in to reply