All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Numbers
Some cubes! (Posted on 2012-08-10) Difficulty: 2 of 5
LIST all the integers equal to the sum of the digits of their cubes.
Prove that your list is exhaustive.

No Solution Yet Submitted by Ady TZIDON    
No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution Both Parts | Comment 4 of 8 |

All you must do is search the first 99 numbers.

Let x be an n digit number. 

It is easy to show that x^3 has between (3n-2) and (3n) digits.

That is, 3n is the maximum digits that x^3 can have.

Now the sum of these digits has a maximum of 9*3*n = 27n.

The minimum value of the number x (an n digit number) is simply 10^(n-1).

It is easy to show that for n>=3:

10^(n-1) > 27n.

Therefore, when x has 3 digits or more, x^3 is always greater than the sum of the digits of x^3.

Therefore one must search just the 2 digit numbers.  I was also able to prove that no number smaller than 54 was possible using the same principle.  (since 99*99*99 is a 6 digit number, and 9*6 = 54.)

The search yields the numbers as 1^3, 8^3, 17^3, 18^3, 26^3 and 27^3.

Edited on August 10, 2012, 3:09 pm
  Posted by Chris, PhD on 2012-08-10 12:56:07

Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (1)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2017 by Animus Pactum Consulting. All rights reserved. Privacy Information