How many nonnegative integers can be written in the form a_7 × 3^7 + a_6 × 3^6 + a_5 × 3^5 + a_4 × 3^4 + a_3 × 3^3 + a_2 × 3^2 + a_1 × 3^1 + a_0 × 3^0, where a_i \in \-1, 0, 1\ for 0 ≤ i ≤ 7?

2018 AMC 12A Problem 13 — Solution

2018 AMC 12A Problem 13

Below is the professionally curated solution for Problem 13 of the 2018 AMC 12A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2018 AMC 12A solutions, or check the answer key.

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Concepts:number base bijection symmetry

Difficulty rating: 1660

13.

How many nonnegative integers can be written in the form

$a_7 \cdot 3^7 + a_6 \cdot 3^6 + a_5 \cdot 3^5 + a_4 \cdot 3^4 + a_3 \cdot 3^3 + a_2 \cdot 3^2 + a_1 \cdot 3^1 + a_0 \cdot 3^0,$

where $a_i \in \{-1, 0, 1\}$ for $0 \le i \le 7?$

$512$

$729$

$1094$

$3281$

$59{,}048$

Solution:

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.

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