+0

0
99
2

What is the remainder when $$1^3+2^3+3^3+\cdots+100^3$$ is divided by 6?

Jun 5, 2020

#1
0

sumfor(n, 1, 100, n^3) = 25502500 mod 6 = 4 - the remainder

Jun 5, 2020
#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:

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



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 $$\boxed{4}$$.

Jun 5, 2020