2018 AMC 10A Problem 18

There are $3^8 = 6561$ choices of the digit string $(a_7, \dots, a_0)$. Two facts settle the count.

All $6561$ sums are distinct. Suppose two different strings gave the same value. Subtracting, we get $\sum (a_i - b_i) 3^i = 0$ with each $a_i - b_i \in \{-2, -1, 0, 1, 2\}$ and not all zero. Look at the smallest index $k$ with $a_k \ne b_k$: every other nonzero term is a multiple of $3^{k+1}$, while $(a_k - b_k)3^k$ is not, so the total cannot be $0$. Contradiction.

The values are symmetric about $0$. Negating every digit negates the sum, so the set of attainable values is closed under $x \mapsto -x$. The largest value is $1 + 3 + \dots + 3^7 = \dfrac{3^8 - 1}{2} = 3280$, so the $6561$ distinct values are exactly $-3280, -3279, \dots, 3279, 3280$ (that is $6561$ consecutive integers, so nothing is skipped).

The nonnegative ones are $0, 1, \dots, 3280$: that is $3281$ integers.

The answer is $\boxed{\textbf{(D)}\ 3281}$.

This is balanced ternary: allowing digit $-1$ instead of $2$ still represents each integer uniquely, because each digit string is a different number just as in ordinary base $3$. A count of "how many values" is then a count of digit strings, and the symmetry between a string and its negative splits those strings evenly around zero, with zero itself counted once: $\dfrac{3^8 + 1}{2}$.

Small cases: with digits $a_0$ only, values $-1, 0, 1$ give $2$ nonnegative; with $a_1, a_0$, values $-4$ to $4$ give $5$; with three digits, $-13$ to $13$ give $14$. The pattern $2, 5, 14 = \dfrac{3^n + 1}{2}$ continues to $\dfrac{3^8 + 1}{2} = 3281$.

• Answering 3^8 = 6561 (all sums, negatives included) or 3^8 / 2 without accounting for zero, or assuming different digit strings can collide.
• Computing the maximum as $3^7 = 2187$ or as $3^8 - 1$ instead of the geometric sum $\dfrac{3^8 - 1}{2} = 3280$.