[Prime]

Ben asked to find a number, which is a cube of the sum of its digits.

Clue ran a program performing some exhaustive search under 1,000,000 and found that the cubes of 0, 1, 8, 17, 18, 26, and 27 have that property.

I claim that I can find more numbers with that property even under 100,000. And I don't need to use a computer for that.

Help my argument.

[Guest]

Ben asked to find a number, which is a cube of the sum of its digits.

Clue ran a program performing some exhaustive search under 1,000,000 and found that the cubes of 0, 1, 8, 17, 18, 26, and 27 have that property.

I claim that I can find more numbers with that property even under 100,000. And I don't need to use a computer for that.

Help my argument.

For clarification, 0,1,8,17,18,26 and 27 are the numbers to be cubed. The sum of the digits of their cubes equal themselves. The "under 1,000,000" means 1,000,000 before cubing. I'm not entirely sure whether Prime's "100,000" means before or after cubing.

I am as interested as he is in this, because I don't think it's possible.

[Guest]

Maybe the problem lies in 0.

I dont think 0^3 should be 0. If I would have to give an answer what 0^3 is I would say 1.

[Guest]

I ran a complete search from 1 to 1,000,000 in which I:

cubed the number,

added the digits in the cube,

and compared the result to the original number.

I found no new answers other than the posted ones.

Perhaps:

the answer involves interpreting the question in a different way? Perhaps not restricting the problem to integers?

[Guest]

One more question:

'Number' means non-negative integer?

Edited August 17, 2008 by epylar

[Guest]

Ben asked to find a number, which is a cube of the sum of its digits.

Clue ran a program performing some exhaustive search under 1,000,000 and found that the cubes of 0, 1, 8, 17, 18, 26, and 27 have that property.

I claim that I can find more numbers with that property even under 100,000. And I don't need to use a computer for that.

Help my argument.

I guess if you use non decimal system other solutions could be found too

[Guest]

I guess if you use non decimal system other solutions could be found too

Verified in bases 2,3,4,5,6 (there are more solutions in these bases).

[Guest]

One more question:

'Number' means non-negative integer?

The integer restriction would be unnecessary, since the sum of digits will always be an integer, and the cube of an integer will also be an integer.

The non-negative restriction may be unnecessary, depending on how you define "sum of digits." For example, would the sum of the digits of -512 be

(a) -5+1+2,

(b) 5+1+2, or

© -(5+1+2)?

If it's c, then the solutions would be -1* the positive solutions. If b then there are no solutions, since a positive number cubed will always be positive. If it's a, the question is a little more interesting. After looking, though, I think this only adds -1 as a solution.

Also OB, 0³ is definitely 0. I think you're getting confused by the fact that x⁰=1. 0³, though, is just 0x0x0=0.

EDIT: On another note, how can I make "©" not come out as a copyright symbol?

Edited August 18, 2008 by Chuck Rampart

[rookie1ja]

EDIT: On another note, how can I make "©" not come out as a copyright symbol?

C) or c) or ( c) ...

[Prime]

The integer restriction would be unnecessary, since the sum of digits will always be an integer, and the cube of an integer will also be an integer.

The non-negative restriction may be unnecessary, depending on how you define "sum of digits." For example, would the sum of the digits of -512 be

(a) -5+1+2,

(b) 5+1+2, or

© -(5+1+2)?

If it's c, then the solutions would be -1* the positive solutions. If b then there are no solutions, since a positive number cubed will always be positive. If it's a, the question is a little more interesting. After looking, though, I think this only adds -1 as a solution.

Also OB, 0³ is definitely 0. I think you're getting confused by the fact that x⁰=1. 0³, though, is just 0x0x0=0.

EDIT: On another note, how can I make "©" not come out as a copyright symbol?

Forget negative. Cosider all digits positive. I should have stipulated that in the OP. It's my omission.

As a side note: If you cannot defeat it -- use it.©Prime.

[Prime]

I guess if you use non decimal system other solutions could be found too

THAT'S A GOOD GUESS!

Here are some examples:

Ben never stipulated, the number must be in decimal. And I feel more comfy with the bases of power 2, anyway. So I never default to decimal. SOME EXAMPLES:

For 8:

2³ = 20 (in base 4.)

or

(1+1)³ = 11 in base 7.

For 27:

(2+1)³ = 21 in base 13

or

(1+2)³ = 12 in base 25.

For 64:

(3+1)³ = 31 in base 21.

or

(1+2+1)³ = 121 in base 7.

And so on. In fact, you can find such an arrangement for any cube under 100,000 in one base system, or another.