AMC 10 · 2017 · #17

Grade 12 algebra

Archetype Modular Cycle Frequency Count └ Sub-archetype Power-Cycle Residue

roots-of-unitycomplex-polar-formparity systematic-enumerationpattern-recognition ↑ Prerequisites: complex-numbersroots-of-unity

📏 Medium solution 💡 3 insights

Problem

Every solution of a root-of-unity equation is raised to a power. Count the real results.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

How to solve

Strategy Organize Information in More Ways

Written as a + bi, the 24 solutions are ugly numbers full of square roots, and raising one of them to the sixth power by hand is hopeless. The same 24 numbers have a second storage format: each is a point on the unit circle, recorded by its angle. In that format multiplying two numbers just adds their angles, so the sixth power is a single multiplication on one integer. So the plan is to re-file each solution as an angle, name that angle with an index variable, translate "z⁶ is real" into a plain condition on the index, and count the indices that satisfy it.

1STEP 1

Name the solutions with an index

An index names every solution.

z_k = cos((π k)/12) + i sin((π k)/12), k = 0, 1, …, 23

2STEP 2

Picture them as a regular 24-gon

They sit evenly around a circle.

2π/24 = π/12 radians = 15° between neighbouring points

3STEP 3

Raise to the sixth: multiply the angle

Raising to a power multiplies the angle.

z_k⁶ = cos((π k)/2) + i sin((π k)/2)

4STEP 4

Translate "real" into one equation

Being real is one sine equation.

z_k⁶ ∈ R ⇔ sin((π k)/2) = 0

5STEP 5

Solve the condition for k

It holds exactly when the index is even.

(π k)/2 = mπ ⇔ k/2 = m ⇔ k is even

6STEP 6

Count the even indices

Counting those gives 12, choice (D).

k ∈ {0, 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22}, count = 24/2 = 12

Answer

Spot-check the two kinds of index. For k = 1 the sixth power has angle π/2, giving z⁶ = i, which is not real — an odd index fails, as claimed. For k = 2 the angle is π, giving z⁶ = -1, which is real — an even index passes. The two extreme choices are also ruled out directly: (A) 0 is impossible because z = 1 obviously works, and (E) 24 is impossible because k = 1 just failed. The tempting wrong answer is (C) 6, which counts only the z with z⁶ = 1 and forgets that z⁶ = -1 is real too; doubling that count gives 12 and matches. A smaller version of the same problem confirms the shape: for z⁸ = 1, the number z² has angle (π k)/2 as well, so again only even k works, which is 4 of the 8 solutions — half, just as here.

💡Key takeaway

Store a complex number as an angle on the circle, and raising it to a power becomes multiplying that angle — then the question is just about whole numbers.

Name the solutions with an index
Picture them as a regular 24-gon
Raise to the sixth: multiply the angle
Translate "real" into one equation
Solve the condition for k
Count the even indices