1. Originally Posted by wonderboy1953
This part is correct: "Finally, an explanation. So the challenge is for computers and/or programmers: what are the numbers such that when cubed the sum of the cube's digits equals a cube?" but note my PS: "PS The number you get after adding up the summands can't be just any cube, but the number you started off with when you were cubing. So if it were true that you got 43 instead of 28, then you have one of the answers which didn't turn out to be the case." So that cube must equal the cube you started off with (and there are six such cubes).

How much progress have you made?

Once again you succeed to confuse me: THEN you can NOT begin with any number but with a cube, since you want

that this number, when cubed (once again) , we'll get:

(1) a cube (again!),

(2) the very same number we began with! (Of course, only condition (2) is required here since

condition (1) follows at once if we begin with a cube, but I'm already wary of all this so I'd better

be repetitive...)

In the body of your explanation you wrote "Now my challenge is to take a number, cube it, and then

add up its digits to see if you get ==>> another cube number <<== from the digit summands" (the

remarked part is mine), but then you say, in your PS, that that sum must be THE same number you began with!

Thus, your example 43 was NOT a good example (nor bad one: not a example at all but a misleading

piece of data) since 43 is not a cube.

Examples (of what you, hopefully, want) would be 0, 1, 8, 27, ...right?

Tonio

2. Originally Posted by tonio
Once again you succeed to confuse me: THEN you can NOT begin with any number but with a cube, since you want

that this number, when cubed (once again) , we'll get:

(1) a cube (again!),

(2) the very same number we began with! (Of course, only condition (2) is required here since

condition (1) follows at once if we begin with a cube, but I'm already wary of all this so I'd better

be repetitive...)

In the body of your explanation you wrote "Now my challenge is to take a number, cube it, and then

add up its digits to see if you get ==>> another cube number <<== from the digit summands" (the

remarked part is mine), but then you say, in your PS, that that sum must be THE same number you began with!

Thus, your example 43 was NOT a good example (nor bad one: not a example at all but a misleading

piece of data) since 43 is not a cube.

Examples (of what you, hopefully, want) would be 0, 1, 8, 27, ...right?

Tonio
As long as 0, 1, 8, 27 are what you started off with.

Here's a different example of what I'm talking about, but this time I'll do it with a square.

Say you're checking the number 9 to see if the sum of the digits to its square is also 9. So 9^2 = 81 and 8 + 1 = 9 which is what you started off with so 9 does fulfill the conditions of the problem (when you're considering squares). Now instead of squares, check the cubes of numbers, sum their digits and see if you can get the original number back that you just cubed.
If you can, then you've found one the six answers to the problem.

Go get 'em Tonio, I know you can do it.

3. Originally Posted by wonderboy1953
As long as 0, 1, 8, 27 are what you started off with.

Here's a different example of what I'm talking about, but this time I'll do it with a square.

Say you're checking the number 9 to see if the sum of the digits to its square is also 9. So 9^2 = 81 and 8 + 1 = 9 which is what you started off with so 9 does fulfill the conditions of the problem (when you're considering squares). Now instead of squares, check the cubes of numbers, sum their digits and see if you can get the original number back that you just cubed.
If you can, then you've found one the six answers to the problem.

Go get 'em Tonio, I know you can do it.

"As long as 0, 1, 8, 27 are what you started of with"...Of course, what else?!

So I've found four non-negative integers that fulfill what you want, and you said there are 6.

Nevertheless , for $\displaystyle n\geq 4$ , the number of digits in $\displaystyle n^6$ is not enough to get, when

added, the number $\displaystyle n^3$ itself again, so could you finally say which ones are the

other 2 numbers besides 0,1,8, 27?

Write them down , Wonderboy, I know you can do it (though I'm not sure what will the outcome be)

Tonio

4. Originally Posted by wonderboy1953
Let's say we take the cube of 43 which is
79507. When you add the digits you get 28. Is 28 a cube? No.

Now my challenge is to take a number, cube it, and then add up its digits to see if you get another cube number from the digit summands. What numbers are these (my understanding is there are only six candidates which are in the OEIS). So get your computers ready to find these numbers.
The way I'm looking at it, I get way more than six answers:

$\displaystyle 1^3 = 1$, these digits add up to $\displaystyle 1$ which is a cube.
$\displaystyle 10^3 = 1000$, these digits add up to $\displaystyle 1$ which is a cube.
$\displaystyle 100^3 = 1000000$, these digits add up to $\displaystyle 1$ which is a cube.

And so on...

Edit: Nevermind...

PS The number you get after adding up the summands can't be just any cube, but the number you started off with when you were cubing. So if it were true that you got 43 instead of 28, then you have one of the answers which didn't turn out to be the case.

5. Code:
int main(){
int i, count = 0, x = 1;
while(count < 6){
int cube = x*x*x;
int len = floor(log(cube)/log(10)+.0001)+1;
int sum = 0;
int pow = 1;
for(i = 0; i < len; i++){
sum += (cube % (10*pow))/pow;
pow *= 10;
}
if(x == sum){
count++;
printf("%i\n",x);
}
x++;
}
return 0;
}
Output: $\displaystyle 1,8,17,18,26,27$

6. Originally Posted by chiph588@
Code:
int main(){
int i, count = 0, x = 1;
while(count < 6){
int cube = x*x*x;
int len = floor(log(cube)/log(10)+.0001)+1;
int sum = 0;
int pow = 1;
for(i = 0; i < len; i++){
sum += (cube % (10*pow))/pow;
pow *= 10;
}
if(x == sum){
count++;
printf("%i\n",x);
}
x++;
}
return 0;
}
Output: $\displaystyle 1,8,17,18,26,27$

I got those same long ago, but according to Wonderboy 17, 18, 26 are not good since they themselves aren't cubes, as he

explained some posts ago.

The only ones I succeed to find are 0, 1, 8 and 27, but he said there are 6...I really want to see those two numbers more not in my list.

Tonio

7. Originally Posted by tonio
I got those same long ago, but according to Wonderboy 17, 18, 26 are not good since they themselves aren't cubes, as he

explained some posts ago.

The only ones I succeed to find are 0, 1, 8 and 27, but he said there are 6...I really want to see those two numbers more not in my list.

Tonio

Soooo?? Where are those two cubes more? There's always the possibility to do spoiler and not to "spoil" the fun for others.

I'm still waiting.

Tonio

8. ## chiph588 is the winner

Since chiph588 is the first to post in the right set of answers (tonio, both you and him had access to the same information on this thread, but he first put up the right set of six answers which makes him the winner).

I want to note that while you could count 0 as an answer, I wouldn't do so since 0 isn't a counting number which number series don't normally count either.

The OEIS number is A061209 on this.

Now I'll give tonio a chance to redeem himself. Can you prove there aren't any further right answers and what is the name given to this set of numbers?

9. Originally Posted by wonderboy1953
Since chiph588 is the first to post in the right set of answers (tonio, both you and him had access to the same information on this thread, but he first put up the right set of six answers which makes him the winner).

I want to note that while you could count 0 as an answer, I wouldn't do so since 0 isn't a counting number which number series don't normally count either.

The OEIS number is A061209 on this.

Now I'll give tonio a chance to redeem himself. Can you prove there aren't any further right answers and what is the name given to this set of numbers?

That is ridiculous because

(1) This is not a contest

(2) You specifically said that the the original number had to be a cube, and 17,18, 26 are not cubes.

(3) The OEIS A061209 is something very different from what you said, from what

Chip said and from what you thought, according to you "explanation, it was.

I suspected something like this almost from the beginning: the question was confusing and ill-posed

and the explanation later was nonsense as well.

I'll give you one more chance to redeem yourself and accept here you were wrong and, perhaps, to stop using that childish,

annoying and patronizing tone when addressing professional mathematicians.

Tonio

Ps. The OEIS A061209 title is simply "Numbers which are the cubes of their digit sum.", and not

the nonsense you wrote in your first post, and of course much less the ""explanation"

you gave after that and by which you confused things even more.

The sequence, as the title says, is something very different: the listed numbers there

are, of course, 1, 512, 4913, 5832, 17576, 19683 , and not what I, Chip, and what you, though it to be from the beginning.

You know, in mathematics one must be direct, clear and as simple as possible. You were neither.

Better luck and much better attitude for the next time.

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