2018 AMC 10B Problem 16

We cannot possibly find the $2018$ numbers, so the answer must depend only on their sum. Look for a link between a number and its cube when dividing by $6$.

1. Try small numbers.

Each cube leaves the same remainder as the number itself.
2. Why it is always true. The difference is
$$n^3 - n = (n-1) \times n \times (n+1),$$
the product of three numbers in a row. Among three numbers in a row, one is even and one is a multiple of $3$. So the product is a multiple of $6$. That means $n^3$ and $n$ always leave the same remainder when divided by $6$.
3. Add them up. Since each cube differs from its number by a multiple of $6$, the sum of the cubes differs from the plain sum by a multiple of $6$. So the sum of cubes has the same remainder as
$$a_1 + a_2 + \cdots + a_{2018} = 2018^{2018}.$$
4. Remainder of $2018^{2018}$. $2018 = 6 \times 336 + 2$, so only the $2$ matters. Powers of $2$ divided by $6$:
- $2 \to 2$, $4 \to 4$, $8 \to 2$, $16 \to 4$, $32 \to 2$, ...

Odd powers leave $2$ and even powers leave $4$. The power $2018$ is even, so the remainder is $4$.

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

When a problem gives far too little information to find the numbers, it is asking about something that does not depend on them. Here that is the fact that a number and its cube always leave the same remainder when divided by $6$. The words "strictly increasing" do not matter at all.

Find the remainder of $2018^{2018}$ by thinking about $2$ and $3$ separately. It is even, so its remainder when divided by $2$ is $0$. Also $2018$ is one less than a multiple of $3$, and an even power of such a number is one more than a multiple of $3$. Among $0, 1, 2, 3, 4, 5$, the only number that is even and one more than a multiple of $3$ is $4$.

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 $\boxed{\textbf{(E)}\ 4}$.