2018 AMC 12A Problem 13

Question

How many nonnegative integers can be written in the form

\begin{aligned} &a_7 \cdot 3^7 + a_6 \cdot 3^6 + a_5 \cdot 3^5 \\ &\quad {}+ a_4 \cdot 3^4 + a_3 \cdot 3^3 + a_2 \cdot 3^2 \\ &\quad {}+ a_1 \cdot 3^1 + a_0 \cdot 3^0, \end{aligned}

where

a_i \in \{-1, 0, 1\}

for

0 \le i \le 7?

512

729

1094

3281

59{,}048

Shifting each

a_i

from

\{-1, 0, 1\}

to

\{0, 1, 2\}

makes an ordinary base-

3

numeral

The

3^8

representable integers are symmetric about

0;

count

0

together with half of the remaining values

Adding

1

to every

a_i

gives a bijection between these expressions and the base-

3

numerals for

0

through

3^8 - 1,

so exactly

3^8 = 6561

distinct integers occur. They are symmetric about

0

(negating all

a_i

negates the value), so besides

0

itself, half are positive:

1 + \tfrac12(6561 - 1) = 3281

nonnegative integers, namely

0

through

3280.

Thus, the correct answer is D.

Problem 13 in other years