Sums of Consecutive Cubes (Trouble Interpreting Question) Show that it is possible to divide the set of the first twelve cubes $\left(1^3,2^3,\ldots,12^3\right)$ into two sets of size six with equal sums.
Any suggestions on what techniques should be used to start the problem?
Also, when the question is phrased like that, are you to find a general case that always satisfies the condition? Or, do they instead want you to find a specific example, since if an example exists then of course the case would be possible. 
Thanks!
Edit: I finally found an answer :) $$1^3 + 2^3 + 4^3 + 8^3 + 9^3 + 12^3 = 3^3 + 5^3 + 6^3 + 7^3 + 10^3 + 11^3 = 3042$$
My approach to solving this problem was an extension of David's suggestion below. For six cubes to sum to an even number (3042), there has to either be 0 odd cubes, 2 odd cubes, 4 odd cubes, or 6 odd cubes. There cannot be 3 odd numbers because the sum of 3 odd numbers results in an odd number. The remaining 3 numbers would be even perfect cubes, and their sum would be even. 
Thus you would have an even + odd = odd sum, but 3042 is even, not odd. Following this logic, a set of six cubes that sum to 3042 can only have 0 odds, 2 odds, 4 odds, or 6 odds (note that the 0 odd case is the complementary set to the 6 odd case, and the 2 odd case is the complementary set to the 4 odd case).
Checking the 0 or 6 odd case is simple. The sum of all the odd cubes does not equal 3042, so the two sets cannot be composed of all odd or no odds.
Hence one set of six cubes must have 2 odds, and the other set must have 4 odds.
Now it is simply a matter of guess and check, checking all the pairs of odd cubes from 
($1^3,3^3$) $\rightarrow$ ($9^3,11^3$). Fortunately, ($1^3,9^3$) works, so we don't have to try too many cases.
Also, if anyone has any other solution method, please let me know :)

Work modulo 7. $x^3=1$ if $x\in A=\{1, 2, 4, 8, 9, 11\}$; $x^3=-1$ if $x\in B=\{3, 5, 6, 10, 12\}$; $x^3=0$ if $x=7$. $3042=4=-3$ mod 7. As barak manos has pointed out, $11$ and $12$ are in different groups. Each group must have either three more elements of $B$ than of $A$, or four more elements of $A$ than of $B$. If the group with $11$ has three more elements of $B$ than of $A$, it has all of $B$ except $12$. And this works. (If the group with $11$ contains four more elements of $A$ than of $B$, it has 5 of $A$ and one of $B$. So the group with $12$ has 4 of $B$, 1 of $A$ and $7$; $3042-12^3-7^3=971$ so it omits $10^3$, but $971-6^3-5^3-3^3=603$ doesn't work.)
