AMC 10 · 2009 · #15

Grade 11 algebra
Archetype Modular Cycle Frequency Count └ Sub-archetype Power-Cycle Residue
complex-numbersperiodic-functionsequences-arithmetic pattern-recognitioncasework ↑ Prerequisites: complex-numbers
📏 Medium solution 💡 3 insights
Problem
A sum weights each power of the imaginary unit by its own position. Find the length that makes the total equal a given complex number.

Pick an answer.

(A)
24
(B)
48
(C)
49
(D)
97
(E)
98
How to solve
Strategy Identify Subproblems

Tool #5 (Look for a Pattern) spots the one fact that makes this sum tractable: i⁴ = 1, so the factors i^k repeat with period 4. That suggests cutting the sum into consecutive blocks of four, which is Tool #7 (Identify Subproblems) — the whole sum becomes one repeated subproblem plus a short leftover. The critical discipline here is to prove the block value rather than read it off the first two blocks: the coefficients 4m+1, 4m+2, 4m+3, 4m+4 change from block to block, so the claim that every block contributes the same amount is an algebraic identity that has to be checked in general, not a pattern. Tool #4 (Introduce a Variable) supplies the general block index m and writes n = 4m + r so that every positive integer n is covered exactly once. Tool #3 (Eliminate Possibilities) then kills three of the four remainder classes on sign grounds alone, and Tool #13 (Convert to Algebra) turns the surviving class into a pair of linear equations in m that must both hold.

1STEP 1

The multipliers repeat every four

The multipliers repeat every four terms.

i¹ = i, i² = -1, i³ = -i, i⁴ = 1, i^k+4 = i^k
2STEP 2

Every block of four sums to 2-2i

Inside each block the weights cancel, leaving a constant.

Σ_j=1⁴(4m+j) i⁴m+j = 4m(i-1-i+1)_= 0 + (i-2-3i+4) = 2-2i
3STEP 3

Exact value of the sum for every n

That gives an exact value for every length.

S(4m)=2m-2mi; S(4m+1)=2m+(2m+1)i; S(4m+2)=-(2m+2)+(2m+1)i; S(4m+3)=-(2m+2)-(2m+2)i
4STEP 4

Three of the four cases die on sign

Signs kill three of the four cases.

n ≡ 2, 3 (mod 4) → Re S(n) ≤ -2; n ≡ 0 (mod 4) → Im S(n) ≤ 0
5STEP 5

Solve the one surviving case

Solving the survivor gives 97, choice (E).

2m = 48 and 2m+1 = 49 → m = 24 → n = 4(24)+1 = 97
Answer
97
A rough size check first: in the surviving case the real part is 2m ≈ n/2, so a real part of 48 points at n near 96, and 97 sits right there. More telling is what the table says about the near misses, since the answer choices are built out of them. At n = 96 the sum is 48 - 48i: the real part is already correct, so any argument that looks only at the real part cannot separate 96 from 97, and 96 is excluded here by the imaginary part rather than by its absence from the option list. At n = 98 the sum is -50 + 49i: now the imaginary part is correct and the real part is not, which is exactly why 98 appears as choice (E). Choices (A) 24, (B) 48, and (C) 49 are the block count and the two components of the target — the quantities a solver holds mid-computation. Every distractor is a genuine intermediate value, and the four-row table disposes of all of them at once. A term-by-term check of the running total for n up to a few hundred confirms 48 + 49i occurs exactly once, at n = 97.
💡Key takeaway

Powers of i repeat every four steps, so cut the sum into blocks of four — each block is exactly 2 - 2i no matter where it starts — and only the leftover terms at the end decide the answer.

  • The multipliers repeat every four
  • Every block of four sums to 2-2i
  • Exact value of the sum for every n
  • Three of the four cases die on sign
  • Solve the one surviving case

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