Question

How many nonnegative integers can be written in the form $a_7\cdot3^7+a_6\cdot3^6+a_5\cdot3^5$ $+a_4\cdot3^4+a_3\cdot3^3+a_2\cdot3^2$ $+a_1\cdot3^1+a_0\cdot3^0,$ where $a_i\in \{-1,0,1\}$ for $0\le i \le 7?$

$512$

$729$

$1094$

$3281$

$59,048$

Answer: D

Concepts:number base symmetry pairing and grouping

Difficulty rating: 1770

Solution:

Note that every number formed by this sum is either positive, negative, or zero.

The number of positive numbers equals the number of negative numbers due to symmetry (flip the $1$ s to $-1$ s and $-1$ s to $1$ s).

The only way for the sum to be $0$ is if all the coefficients are $0.$

The total number of numbers is $3^8 = 6561.$ Because each power of $3$ is larger than the sum of all previous powers of three, each combination of coefficients yields a different value. More explicitly, at the highest place where two combinations differ, the difference has magnitude at least $3^k,$ while all lower places together can cancel at most $2(1+3+\cdots+3^{k-1})=3^k-1.$

Therefore, there are $\dfrac{6561 - 1}{2} + 1 = 3281$ distinct nonnegative integers.

Thus, D is the correct answer.

2018 AMC 10A Problem 18

Answer: D

Difficulty rating: 1770

Solution:

Problem 18 in Other Years