2018 AMC 10B Problem 16

The sequence itself is nowhere near determined, so the answer must depend only on the given sum. The bridge between a number and its cube modulo $6$ is the identity
$$n^3 - n = (n-1)\,n\,(n+1),$$
a product of three consecutive integers, which always contains a multiple of $2$ and a multiple of $3$. Hence $n^3 \equiv n \pmod 6$ for every integer $n$.

Applying this to each term and adding,
$$a_1^3 + \cdots + a_{2018}^3 \equiv a_1 + \cdots + a_{2018} = 2018^{2018} \pmod 6.$$

Now $2018 = 6 \cdot 336 + 2$, so $2018^{2018} \equiv 2^{2018} \pmod 6$. Powers of $2$ modulo $6$ go $2, 4, 2, 4, \dots$: odd exponents give $2$, even exponents give $4$. Since $2018$ is even, $2^{2018} \equiv 4 \pmod 6$.

The answer is $\boxed{\textbf{(E)}\ 4}$.

When a problem gives far too little information to find the objects, it is asking for an invariant. Here the invariant is $n^3 \equiv n \pmod 6$ (equivalently, cubing preserves the residue mod $2$ and mod $3$ by Fermat's little theorem). The "strictly increasing" condition is pure decoration.

Check $n^3 \equiv n \pmod 6$ by listing residues: $0^3, 1^3, \dots, 5^3 = 0, 1, 8, 27, 64, 125 \equiv 0, 1, 2, 3, 4, 5$. Then reduce $2018^{2018}$ separately mod $2$ (it is $0$) and mod $3$ ($2018 \equiv 2 \equiv -1$, and $(-1)^{2018} = 1$); the residue that is even and $1 \bmod 3$ is $4$.

• Trying to determine the actual terms (the sequence is not determined), or reducing 2018^2018 mod 6 as 2^2018 = 2 instead of 4 (even powers of 2 are 4 mod 6).
• Guessing $0$ because "the sum is huge and even" or because cubes feel divisible by $6$; only the residue of the sum matters, and $2018^{2018}$ is $4 \bmod 6$, not $0$.