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 Some cubes! (Posted on 2012-08-10)
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

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 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

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