LIST all the integers equal to the sum of the digits of their cubes.

Prove that your list is exhaustive.

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**