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
4 years, 8 months ago

No vote yet
1 vote

  Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

  • Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
  • Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
  • Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
  • Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 2×3 2 \times 3
2^{34} 234 2^{34}
a_{i-1} ai1 a_{i-1}
\frac{2}{3} 23 \frac{2}{3}
\sqrt{2} 2 \sqrt{2}
\sum_{i=1}^3 i=13 \sum_{i=1}^3
\sin \theta sinθ \sin \theta
\boxed{123} 123 \boxed{123}

Comments

Sort by:

Top Newest

No output in 4 hours!! :(

Pranjal Jain - 4 years, 8 months ago

Log in to reply

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

Soutrik Bandyopadhyay - 4 years, 8 months ago

Log in to reply

Great question to ask!

Calvin Lin Staff - 4 years, 8 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 - 4 years, 8 months ago

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

David Holcer - 4 years, 8 months ago

Log in to reply

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

Ronak Agarwal - 4 years, 8 months ago

Log in to reply

@Ronak Agarwal Depends on programming language you are working upon

Pranjal Jain - 4 years, 8 months ago

Log in to reply

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

David Holcer - 4 years, 8 months ago

Log in to reply

@David Holcer Probably there exists some large xx for which the inequality is satisfied. Probably finite solutions exist only when you consider xkx^{k} for very very large kk.

Jake Lai - 4 years, 8 months ago

Log in to reply

@Jake Lai Clearly if there is one such xx then there are infinitely many such xx; just append zeroes at the end.

Ivan Koswara - 4 years, 8 months ago

Log in to reply

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

Ronak Agarwal - 4 years, 8 months ago

Log in to reply

@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 - 4 years, 8 months ago

Log in to reply

@David Holcer Not till now

Pranjal Jain - 4 years, 8 months ago

Log in to reply

@Pranjal Jain What, what number?!!

David Holcer - 4 years, 8 months ago

Log in to reply

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

Pranjal Jain - 4 years, 8 months ago

Log in to reply

@Pranjal Jain Oh, ok.Thnx.

David Holcer - 4 years, 8 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...