#2**0 **

Let's use modulos. If you don't know what those are, they're basically the remainder when you divide something by something else, for example \(8 \equiv 2 \pmod 6.\) We can use these here so that we don't need to continually repeat, and only need to check \(1\) to \(6 \pmod 6\). (Technically it should start at \(0\), but in this problem this is easier since it starts at \(1 \pmod 6\).

We can take the cube of each and we find:

\(1^3 \equiv 1 \pmod 6\\ 2^3 \equiv 8 \pmod 6 \equiv 2 \pmod 6 \\ 3^3 \equiv 27 \pmod 6 \equiv 3 \pmod 6\\ 4^3 \equiv 64 \pmod 6 \equiv 4 \pmod 6 \\ 5^3 \equiv 125 \pmod 6 \equiv 5 \pmod 6 \\ 6^3 \equiv 216 \pmod 6 \equiv 0 \pmod 6.\)

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When we add all of them up, we get \(3 \pmod 6\), which means for every 6 numbers they sum to \(3 \pmod 6\). Thus, from \(1\) to \(96\), every \(6\) sum to \(3 \pmod 6\), and there are \(16\) sets, so they sum to \(48 \pmod 6 \equiv 0 \pmod 6.\)

Thus, those all cancel out and we only care about \(97,98,99,100\) whose cubes we find are \(1,2,3,4 \pmod 6\), respectively. These add up to \(10 \pmod 6 \equiv 4 \pmod 6\), so the remainder when dividing by 6 is \(\boxed{4}\).

Mathcounts201998 Jun 5, 2020