Expressing $2002^{2002}$ as a sum of cubes

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!!

$k=4$ 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 $4 \pmod 9$.