AMC 10 · 2018 · #18

Grade 7 arithmetic

Archetype Overlap / Complement Counting └ Sub-archetype Complement by Subtracting Forbidden Arrangements

base-conversionexponents symmetry-argumentcomplementary-counting ↑ Prerequisites: base-conversion

📏 Medium solution 💡 2 insights

Problem

Each of the eight coefficients a₇, a₆, …, a₀ is chosen from {-1, 0, 1}, and they build the number a₇· 3⁷ + a₆· 3⁶ + … + a₁· 3¹ + a₀· 3⁰. Count how many different nonnegative integers (zero or positive) can come out.

Pick an answer.

(A)

512

(B)

729

(C)

1094

(D)

3281

(E)

59,048

AMC 10 2018 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Change Focus / Count the Complement

Tool #2 (Systematic List) counts the raw number of coefficient tuples: 3 choices per digit, 8 digits, so 3⁸ tuples. Tool #14 (Extreme Principle) proves each tuple lands on a different integer by bounding the lower digits against the leading one. Tool #16 (Change Focus) is the key move: instead of counting nonnegative values head-on, use the sign-flip symmetry — positives and negatives come in equal numbers, with exactly one zero, so the nonnegative count is just (total + 1) / 2. Tool #3 (Eliminate) confirms the matching choice and rules out the decoys.

1STEP 1

Count all coefficient tuples

Eight digits, three independent choices each, so by the counting principle there are 3⁸ = 6561 coefficient tuples.

3⁸ = 6561 coefficient tuples

2STEP 2

Each tuple gives a distinct integer

The leading nonzero digit outweighs every lower one, so no two tuples collide — 6561 different integers (this is balanced ternary).

3⁰ + 3¹ + … + 3^k-1 = (3^k - 1)/2 ＜ 3^k

3STEP 3

Use sign-flip symmetry

Flipping every sign sends v to -v, so positives and negatives occur in equal numbers; only the all-zero tuple gives 0.

a_i → -a_i sends v → -v, so #positive = #negative

4STEP 4

Solve for the nonnegative count

With P positives, N negatives and one zero: P + N + 1 = 6561 and P = N give P = 3280, so the nonnegative count is P + 1 = 3281.

2P + 1 = 6561 → P = 3280, P + 1 = 3281

5STEP 5

Match the answer choice

The nonnegative count 3281 is choice (D); the decoys 59,048, 729, 512, and 1094 are wrong-sized powers or fractions.

3281 → (D)

Answer

3281

The range argument and the symmetry agree. The largest value is all coefficients +1: 3⁷ + … + 3⁰ = (3⁸ - 1)/2 = 3280, and the smallest is -3280. Since every tuple is distinct, the expression hits every integer from -3280 to 3280 — that is 6561 integers, matching 3⁸. The nonnegative ones run 0, 1, 2, …, 3280, which is 3281 numbers, confirming (D).

💡Key takeaway

There are 3⁸ = 6561 ways to pick the digits, each gives a different number from -3280 to 3280, and since positives and negatives match up evenly with one zero in the middle, the nonnegative ones number (6561 + 1)/2 = 3281.

Count all coefficient tuples
Each tuple gives a distinct integer
Use sign-flip symmetry
Solve for the nonnegative count
Match the answer choice

Count all coefficient tuples

Each tuple gives a distinct integer

Use sign-flip symmetry

Solve for the nonnegative count

Match the answer choice

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