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This is the problem: Determine the smallest positive integer $k$ such that there exist integers $x_1, x_2 , \ldots , x_k$ with ${x_1}^3+{x_2}^3+{x_3}^3+\cdots+{x_k}^3=2002^{2002} $. How to approach these kind of problems?? Thanks in advance!!

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  • $\begingroup$ Well, since $2002^{2001}$ is a cube number, certainly $k = 2002$ is possible, and I'm sure it's possible to do better... $\endgroup$ – Ben Millwood Jun 26 '12 at 12:40
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    $\begingroup$ @BenMillwood You can get $k=4$ with $x_1 = x_2 = 2002^{667}$ and $x_3 = x_4 = 10*2002^{667}$. I suspect this is the smallest possibility but I don't have a proof. $\endgroup$ – Cocopuffs Jun 26 '12 at 12:44
  • $\begingroup$ @Cocopuffs: ah, that is a much more intelligent way to start... I suppose part of the problem is that the $x_i$ need not be positive? $\endgroup$ – Ben Millwood Jun 26 '12 at 12:47
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    $\begingroup$ The smallest positive integer is $1$ </only read title> $\endgroup$ – Neal Jun 26 '12 at 13:17
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    $\begingroup$ Where did you get the problem? $\endgroup$ – Colonel Panic Jun 26 '12 at 15:16
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$k=4$ is the smallest:

Certainly, it can be done using 4 cubes, by noticing that $2002 = 10^3 + 10^3 + 1^3 +1^3$, and then using $2002^{2002} = 2002 \times 2002^{2001} = (10^3 + 10^3 + 1^3 +1^3)\times (2002^{667})^3$, and multiplying out the brackets.

Since the number can be represented by 4 cubes, it suffices to show that it cannot be done with less than 4.

Since $2002 \equiv 4 \pmod 9$ we have $2002^3 \equiv 64 \equiv 1 \pmod 9$ so that $2002^{3n} \equiv 1 \pmod 9$ and so the original number is equivalent to 4 (mod 9).

Looking at cubes mod 9, they are equivalent to 0, 1 or -1, so at least 4 are required for any number equivalent to $4 \pmod 9$.

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  • $\begingroup$ +1. But how did you prove $k=4$ is the smallest? $\endgroup$ – N.S.JOHN Apr 28 '16 at 2:44
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    $\begingroup$ The very last sentence of the answer proves that $4$ is the smallest: it is impossible to get the correct residue (mod 9) with 3 or less. $\endgroup$ – Old John Apr 28 '16 at 6:05
  • $\begingroup$ Thank you. I am new to the concept of modular arithmetic $\endgroup$ – N.S.JOHN Sep 24 '16 at 3:40

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