# Cubes upon cubes

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• Nov 20th 2010, 06:26 AM
wonderboy1953
Cubes upon cubes
Can you name off the cubes whose base number exponents have their digits sum up to the cube of the base number? (yes undefined, this is in OEIS)

Moderator edit: Approved challenge question.
• Nov 20th 2010, 07:10 AM
tonio
Quote:

Originally Posted by wonderboy1953
Can you name off the cubes whose base number exponents have their digits sum up to the cube of the base number? (yes undefined, this is in OEIS)

It must be my three working neurons took some days (quite a few, in fact) off: I don't understand the given data:

"Cubes" (I suppose integer numbers which are cubes of other integers, like 8, 27, 64 and etc.), "whose base numbers

exponents"...what exponents? Of whom or what? For example, with $\displaystyle 64=4^3$ , are we talking of 3 (the exponent of 4)? That seems

to be a triviality, since all the numbers are cubes.

Tonio
• Nov 20th 2010, 07:18 AM
wonderboy1953
Quote:

Originally Posted by tonio
It must be my three working neurons took some days (quite a few, in fact) off: I don't understand the given data:

"Cubes" (I suppose integer numbers which are cubes of other integers, like 8, 27, 64 and etc.), "whose base numbers

exponents"...what exponents? Of whom or what? For example, with $\displaystyle 64=4^3$ , are we talking of 3 (the exponent of 4)? That seems

to be a triviality, since all the numbers are cubes.

Tonio

Try again Tonio (I submitted an answer key to Mr. Fantastic).
• Nov 20th 2010, 11:27 AM
tonio
Quote:

Originally Posted by wonderboy1953
Try again Tonio (I submitted an answer key to Mr. Fantastic).

No, I won't. I try if I understand the question.

This is my last post in this thread unless there's some explanation about the question itself, not its solution.

Tonio
• Nov 20th 2010, 11:53 AM
Bruno J.
I agree with tonio, this question is very badly worded. Perhaps you should give an example of whatever it is you mean.
• Nov 20th 2010, 08:30 PM
Soroban
Hello, wonderboy1953!

Since it is so obvious, maybe Mr. Fantastic can explain it to us.

Quote:

Can you name off the cubes whose base number exponents
have their digits sum up to the cube of the base number?

Here's my interpretation . . .

We have a cube, say, $\displaystyle 3^3 = 27.$

Write 27 in base-two: .$\displaystyle 27 \;=\;11011_2$

. . That is: .$\displaystyle 27 \;=\;2^4 + 2^3 + 2^1 + 2^0$

The "base number exponents" are: .$\displaystyle 4, 3, 1, 0$

And these digits sum up to 8, the cube of the base number, 2.

We have a cube, say, $\displaystyle 16^3 = 4096$

Write 4096 in base-three: .$\displaystyle 4096 \:=\:12,\!121,\!201_3$

. . That is: .$\displaystyle 4096 \;=\;1\!\cdot\!3^7 + 2\!\cdot\!3^6 + 1\!\cdot\!3^5 + 2\!\cdot\!3^4 + 1\!\cdot\!3^3 + 2\!\cdot\!3^2 + 1\!\cdot\!3^0$

The "base number exponents" are: .$\displaystyle 7, 6, 5, 4, 3, 2, 0$

And these digits sum up to 27, the cube of the base number, 3.

So far, I've found two of them . . . right?

• Nov 21st 2010, 08:42 AM
wonderboy1953
Quote:

Originally Posted by Soroban
Hello, wonderboy1953!

Since it is so obvious, maybe Mr. Fantastic can explain it to us.

Here's my interpretation . . .

We have a cube, say, $\displaystyle 3^3 = 27.$

Write 27 in base-two: .$\displaystyle 27 \;=\;11011_2$

. . That is: .$\displaystyle 27 \;=\;2^4 + 2^3 + 2^1 + 2^0$

The "base number exponents" are: .$\displaystyle 4, 3, 1, 0$

And these digits sum up to 8, the cube of the base number, 2.

We have a cube, say, $\displaystyle 16^3 = 4096$

Write 4096 in base-three: .$\displaystyle 4096 \:=\:12,\!121,\!201_3$

. . That is: .$\displaystyle 4096 \;=\;1\!\cdot\!3^7 + 2\!\cdot\!3^6 + 1\!\cdot\!3^5 + 2\!\cdot\!3^4 + 1\!\cdot\!3^3 + 2\!\cdot\!3^2 + 1\!\cdot\!3^0$

The "base number exponents" are: .$\displaystyle 7, 6, 5, 4, 3, 2, 0$

And these digits sum up to 27, the cube of the base number, 3.

So far, I've found two of them . . . right?

Good try Soroban

All of the answers are in base 10 (and there are six answers).

I give Mr. Fantastic permission to restate the problem in better wording to clarify.

Somebody with a computer should be able to work it out in an hour.

Good luck to everybody.

Wonderboy1953
• Nov 21st 2010, 02:49 PM
Bruno J.
Quote:

Originally Posted by wonderboy1953
Good try Soroban

All of the answers are in base 10 (and there are six answers).

I give Mr. Fantastic permission to restate the problem in better wording to clarify.

Somebody with a computer should be able to work it out in an hour.

Good luck to everybody.

Wonderboy1953

By "good try", do you mean "good try trying to understand the question"?
• Nov 21st 2010, 03:15 PM
Also sprach Zarathustra
Quote:

Originally Posted by Bruno J.
By "good try", do you mean "good try trying to understand the question"?

LOL!
• Nov 21st 2010, 03:28 PM
tonio
Quote:

Originally Posted by Bruno J.
By "good try", do you mean "good try trying to understand the question"?

Ah! So the challenge problem here was to understand what was being asked....man, did I fail or what!(Giggle)

Tonio
• Nov 21st 2010, 03:30 PM
Also sprach Zarathustra
Quote:

Originally Posted by tonio
Ah! So the challenge problem here was to understand what was being asked....man, did I fail or what!(Giggle)

Tonio

LOL! Again!
• Nov 22nd 2010, 07:44 AM
wonderboy1953
Quote:

Originally Posted by Bruno J.
By "good try", do you mean "good try trying to understand the question"?

Here's what I mean. I didn't specify what base the answers are to be expressed in (even though normally base 10 is assumed).
So technically Soroban is within his rights and he gave it a good try to meet the challenge.

I figured out a way to make my challenge clear by using a counterexample. 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.

(btw a secondary challenge would be to prove there are only six numbers that fill the bill and my last question is what are these numbers called).

Again good luck everyone.

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.
• Nov 22nd 2010, 08:44 AM
tonio
Quote:

Originally Posted by wonderboy1953
Here's what I mean. I didn't specify what base the answers are to be expressed in (even though normally base 10 is assumed).
So technically Soroban is within his rights and he gave it a good try to meet the challenge.

I figured out a way to make my challenge clear by using a counterexample. 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.

(btw a secondary challenge would be to prove there are only six numbers that fill the bill and my last question is what are these numbers called).

Again good luck everyone.

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.

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?

For example, 0, 1, 2, 5, 8, 10, 100, 1000, 10000, 100000, .....I see infinite numbers like these.

What had exponents and "sum up to the cube of the base number" to do with the above could be another challenge in the "semantics and weird wording" section, imho.

Tonio
• Nov 24th 2010, 03:41 AM
tonio
Quote:

Originally Posted by tonio
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?

For example, 0, 1, 2, 5, 8, 10, 100, 1000, 10000, 100000, .....I see infinite numbers like these.

What had exponents and "sum up to the cube of the base number" to do with the above could be another challenge in the "semantics and weird wording" section, imho.

Tonio

So...?? Is the above correct from the OP's point of view or did he/she mean, after all, something else?

Tonio
• Nov 24th 2010, 07:55 AM
wonderboy1953
Quote:

Originally Posted by tonio
So...?? Is the above correct from the OP's point of view or did he/she mean, after all, something else?

Tonio

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?
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