AMC 10 · 2006 · #18

Grade 6 arithmetic

Archetype Modular Cycle Frequency Count └ Sub-archetype Position-in-Cycle Lookup

recursive-sequencepattern-recognitionmodular-arithmetic identify-subproblems ↑ Prerequisites: recursive-sequence

📏 Medium solution 💡 2 insights

Problem

A sequence a₁, a₂, a₃, … starts with a₁ = 2 and a₂ = 3. From the third term on, each term equals the previous term divided by the term before that:

a_n = \frac{a_{n-1}}{a_{n-2}}

for every n ≥ 3. Find the value of the 2006th term a₂₀₀₆.

Pick an answer.

(A)

$\frac{1}{2}$

(B)

$\frac{2}{3}$

(C)

$\frac{3}{2}$

(D)

(E)

AMC 10 2006 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Look for a Pattern

The term number 2006 is far too large to reach by computing one term at a time. But the rule only looks back two steps, so once a pair of consecutive terms repeats, the whole sequence must repeat from there. The plan is to list the first several terms, watch for the repeat, and then use the repeat length to jump straight to term 2006.

1STEP 1

List the first several terms

Run the rule term by term: dividing by a fraction means multiplying by its reciprocal.

a₃ = a₂/a₁ = 3/2, a₄ = a₃/a₂ = 3/2/3 = 1/2, a₅ = a₄/a₃ = 1/2/3/2 = 1/3, a₆ = a₅/a₄ = 1/3/1/2 = 2/3

2STEP 2

Spot the repeat

Two more terms give a₇ = 2, a₈ = 3 — the starting pair. The rule sees only the last two terms, so the sequence repeats every 6 terms.

a₇ = a₆/a₅ = 2/3/1/3 = 2 = a₁, a₈ = a₇/a₆ = 2/2/3 = 3 = a₂

3STEP 3

Jump to term 2006 with the cycle

Since $2006 = 6 \times 334 + 2$ , term 2006 sits where term 2 does in the block, so a₂₀₀₆ = a₂ = 3, choice (E).

2006 = 6 × 334 + 2 → a₂₀₀₆ = a₂ = 3

Answer

Check the cycle claim directly: the six listed values 2, 3, 3/2, 1/2, 1/3, 2/3 do return to 2, 3 at positions 7 and 8, so the period of 6 is real. The remainder step is easy to test on small term numbers: a₈ should equal a₂ because 8 = 6 + 2, and indeed a₈ = 3. The same logic gives a₂₀₀₆ = a₂ = 3, so the answer sits inside the given choices as expected.

💡Key takeaway

When a sequence starts repeating, find how long the loop is, then use the remainder to see where a far-away term lands in the loop.

List the first several terms
Spot the repeat
Jump to term 2006 with the cycle

List the first several terms

Spot the repeat

Jump to term 2006 with the cycle

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