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Math Help - Cubes upon cubes

  1. #1
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    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.
    Last edited by mr fantastic; November 20th 2010 at 11:33 AM.
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    Quote Originally Posted by wonderboy1953 View Post
    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  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
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    Quote Originally Posted by tonio View Post
    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  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).
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    Quote Originally Posted by wonderboy1953 View Post
    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
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    MHF Contributor Bruno J.'s Avatar
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    I agree with tonio, this question is very badly worded. Perhaps you should give an example of whatever it is you mean.
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    Hello, wonderboy1953!

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


    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, 3^3 = 27.

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

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

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

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


    We have a cube, say, 16^3 = 4096

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

    . . That is: . 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: . 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?

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    Quote Originally Posted by Soroban View Post
    Hello, wonderboy1953!

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



    Here's my interpretation . . .


    We have a cube, say, 3^3 = 27.

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

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

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

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


    We have a cube, say, 16^3 = 4096

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

    . . That is: . 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: . 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
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    MHF Contributor Bruno J.'s Avatar
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    Quote Originally Posted by wonderboy1953 View Post
    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"?
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    MHF Contributor Also sprach Zarathustra's Avatar
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    Quote Originally Posted by Bruno J. View Post
    By "good try", do you mean "good try trying to understand the question"?
    LOL!
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    Quote Originally Posted by Bruno J. View Post
    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!

    Tonio
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    MHF Contributor Also sprach Zarathustra's Avatar
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    Quote Originally Posted by tonio View Post
    Ah! So the challenge problem here was to understand what was being asked....man, did I fail or what!

    Tonio

    LOL! Again!
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    Quote Originally Posted by Bruno J. View Post
    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.
    Last edited by wonderboy1953; November 22nd 2010 at 08:11 AM. Reason: correction
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    Quote Originally Posted by wonderboy1953 View Post
    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
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    Quote Originally Posted by tonio View Post
    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
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    Quote Originally Posted by tonio View Post
    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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