AMC 10 · 2024 · #23

Grade 5 arithmetic

Archetype Arithmetic Expression Decomposition └ Sub-archetype Multi-Segment Path or Count Sum

recursive-sequencesequences-geometricpattern-recognitionlucas-numbers pattern-recognitioneasier-related-problemidentify-subproblems ↑ Prerequisites: recursive-sequencesequences-arithmeticpattern-recognition

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Problem

The Fibonacci sequence is F₁ = 1, F₂ = 1, F_n = F_n-1 + F_n-2. Compute the sum

\frac{F₂}{F₁}

\frac{F₄}{F₂}

\frac{F₆}{F₃}

+ … +

\frac{F₂0}{F₁0}

— that is, Σ_n=1¹⁰

\frac{F₂n}{F_n}

Pick an answer.

(A)

318

(B)

319

(C)

320

(D)

321

(E)

322

AMC 10 2024 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Make a Systematic List

The problem hands us a finite, well-defined sum of 10 ratios. Tool #2 (Make a Systematic List) is the workhorse: list F₁, …, F₂0 in order, then list the 10 ratios. Tool #9 (Solve an Easier Related Problem) gets us moving even before that — compute the first three or four ratios by hand and notice they are integers (1, 3, 4, 7, …). Tool #5 (Look for a Pattern) then turns those into a Fibonacci-style recurrence (a_n = a_n-1 + a_n-2 starting 1, 3), so the rest of the list extends by addition rather than by dividing big Fibonacci numbers. With 10 integers in hand, the answer is one careful sum.

1STEP 1

List Fibonacci numbers F₁ through F₂0 in order. Each one is the sum of the previous two.

F₁, F₂, …, F₂0 = 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765

2STEP 2

Compute the first few ratios a_n = $\frac{F₂n}{F_n}$ by hand: they come out to whole numbers 1, 3, 4, 7, 11.

a₁ = 1, a₂ = 3, a₃ = 4, a₄ = 7, a₅ = 11

3STEP 3

Spot the pattern: 4 = 1 + 3, 7 = 3 + 4, 11 = 4 + 7 — a_n uses the same add-the-last-two rule as Fibonacci, starting 1, 3.

a_n = a_n-1 + a_n-2 for n ≥ 3, a₁ = 1, a₂ = 3

4STEP 4

Extend by addition for the last five ratios: 18, 29, 47, 76, 123 (check: $\frac{F₂0}{F₁0}$ = $\frac{6765}{55}$ = 123).

a₆, a₇, a₈, a₉, a₁0 = 18, 29, 47, 76, 123

5STEP 5

Add the ten ratios: 1 + 3 + 4 + 7 + 11 + 18 + 29 + 47 + 76 + 123 = 319, choice (B).

S = 1 + 3 + 4 + 7 + 11 + 18 + 29 + 47 + 76 + 123 = 319 → (B)

Answer

319

Two sanity checks. (1) Magnitude — the largest term is a₁0 = 123 and the second-largest is a₉ = 76, so the sum is at least 123 + 76 = 199 and at most 10 · 123 = 1230; landing near 319 (just above 123 · 2 = 246, well below 123 · 4 = 492) is sensible. (2) Recurrence reflection — sums of consecutive Lucas-style terms have a tidy closed form: a₁ + a₂ + … + a₁0 = a₁2 - 3. Computing a₁1 = 76 + 123 = 199 and a₁2 = 123 + 199 = 322 gives 322 - 3 = 319, which matches and uses the same recurrence as a free verification. The nearby distractors 318, 320, 321, 322 are exactly the off-by-one/off-by-two arithmetic traps this re-derivation rules out.

💡Key takeaway

Hard-looking Fibonacci ratios become a simple pattern game once you compute the first few: 1, 3, 4, 7, 11 extends by addition to 123, and ten numbers sum to 319 — just Grade 5 division and Grade 4 number-pattern rules.

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