AMC 10 · 2010 · #15

Grade 11 countingalgebra

Archetype Casework by Position / Vertex Type └ Sub-archetype Vertex or Element Role

complex-numbersperiodic-functionfundamental-counting-principle caseworksystematic-enumeration ↑ Prerequisites: complex-numbers

📏 Long solution 💡 3 insights

Problem

Three quantities built from one range must take exactly two distinct values. Count the triples.

Pick an answer.

(A)

149

(B)

205

(C)

215

(D)

225

(E)

235

How to solve

Strategy Make a Systematic List

The condition is about which of three things match, so Tool #2 (Make a Systematic List) organizes it into the three mutually exclusive match patterns and counts each one. Before counting, Tool #3 (Eliminate Possibilities) does the heavy lifting: comparing absolute values shows i^x and (1+i)^y can only ever agree in one place, and a reality check on (1+i)^y pins down exactly which powers can equal an integer z. Tool #5 (Look for a Pattern) supplies the two cycles that make those eliminations possible — i^x repeats every 4 steps, and (1+i)² = 2i generates every power of 1+i. Tool #16 (Change Focus / Count the Complement) then gives an independent recount in the review, counting each pairwise match separately and correcting for the all-equal overlap.

1STEP 1

Turn the condition into three cases

The condition splits into three matching cases.

{A,B,C} has exactly 2 elements ⇔ exactly one of A = B, A = C, B = C holds

2STEP 2

Describe each of the three slots

Each slot ranges over a short, known list.

i^x ∈ {1, i, -1, -i} (each value hit 5 times), (1+i)² = 2i, z ∈ {0, 1, …, 19}

3STEP 3

Size kills the first cross-match

Size alone kills most cross-matches.

|i^x| = 1, |(1+i)^y| = (√(2))^ y → i^x = (1+i)^y ⇔ y = 0 and i^x = 1

4STEP 4

Which powers are integers under 20

Only two powers land in the allowed range.

(1+i)⁴ⁿ = 4ⁿ(-1)ⁿ: 1, -4, 16, -64, 256 → (y,z) = (0,1) or (8,16)

5STEP 5

Count the two cases pinned at 1

The two pinned cases each give 95.

A = B ≠ C: 5 × 1 × 19 = 95; A = C ≠ B: 5 × 1 × 19 = 95

6STEP 6

Count the last case and add

Adding the last case gives 225, choice (C).

B = C ≠ A: 15 + 20 = 35; total = 95 + 95 + 35 = 225

Answer

225

225 out of 20³ = 8000 triples is under 3%, which fits the situation: a value of size 1, a value that keeps growing, and a small integer rarely coincide. An independent recount confirms the number. Count each pairwise match on its own over all 8000 triples: A = B holds for 5 × 1 × 20 = 100 triples, A = C for 5 × 20 × 1 = 100, and B = C for 2 × 20 = 40 (the two pairs (y,z) = (0,1) and (8,16), with x free). Triples where all three agree need A = B = 1 and z = 1, so there are 5 of them, and each was counted in all three of those pair counts. A triple with exactly one matching pair is counted once, so the number with exactly two distinct elements is 100 + 100 + 40 - 3 × 5 = 225, matching the case-by-case total. It is also worth noting that the choice 205 is precisely what a solver gets by finding (1+i)⁰ = 1 but missing (1+i)⁸ = 16, which confirms that the 16 branch is genuinely required rather than decorative.

💡Key takeaway

Compare sizes first: |i^x| is always 1 while |(1+i)^y| keeps growing, so the three values can only coincide in a few spots — find those spots, then count.

Turn the condition into three cases
Describe each of the three slots
Size kills the first cross-match
Which powers are integers under 20
Count the two cases pinned at 1
Count the last case and add