AMC 10 · 2019 · #9

Grade 8 number-theory

Archetype Arithmetic Expression Decomposition └ Sub-archetype Successive Multiplicative Steps

recursive-sequencesequences-arithmeticpattern-recognitiontelescoping-sum pattern-recognitionidentify-subproblems ↑ Prerequisites: recursive-sequencesequences-arithmetic

📏 Medium solution 💡 3 insights

Problem

A sequence starts with 1 and three sevenths, and from the third term on each term is the product of the two before it divided by twice the earlier one minus the later one. The 2019th term is a fraction in lowest terms. Find the sum of its numerator and denominator.

Pick an answer.

(A)

2020

(B)

4039

(C)

6057

(D)

6061

(E)

8078

How to solve

Strategy Organize Information in More Ways

Tool #15 (Reorganize): take reciprocals — set b_n = 1/a_n. The messy multiplicative recursion turns into the clean linear one b_n = 2b_n-1 - b_n-2, i.e., the differences b_n - b_n-1 are constant — b is arithmetic. Tool #5 (Pattern) + Tool #9 (Easier): verify with a₃, a₄ by hand so the arithmetic pattern in b is visible. Then b₂₀₁₉ follows from an arithmetic-sequence formula, and the answer is p + q. Tool #3 matches the result to the five choices.

1STEP 1

Take reciprocals

Taking reciprocals simplifies the rule.

b_n = 1/a_n → b_n = 2 b_n-1 - b_n-2

2STEP 2

Spot the arithmetic sequence

The gap between neighbours is constant.

b_n - b_n-1 = b_n-1 - b_n-2 → {b_n} arithmetic

3STEP 3

Write the general term

First term and gap give the general term.

b₁ = 1, b₂ = 7/3, d = 4/3 → b_n = (4n - 1)/3

4STEP 4

Check a small term

The third term checks out.

a₃ = 3/7/11/7 = 3/11 = 1/b₃ ✓

5STEP 5

Reach the 2019th term

Plug the index into the formula.

b₂₀₁₉ = 8075/3, a₂₀₁₉ = 3/8075

6STEP 6

Confirm it is reduced

Numerator and denominator are coprime.

8075 = 5² · 17 · 19, gcd(3, 8075) = 1 → p = 3, q = 8075

7STEP 7

Add the two

Adding gives 8078.

p + q = 3 + 8075 = 8078

8STEP 8

Match the choice

It matches a choice exactly.

8078 → (E)

Answer

8078

The answer choices give a sharp clue: 8078 = 2 · 4039 ≈ 4 · 2019. That matches q = 8075, since b₂₀₁₉ = (4 · 2019 - 1)/3 has numerator 8075 ≈ 4 · 2019. Also a_n → 0 as n grows (since b_n → ∞), which is consistent with 3/8075 being tiny.

💡Key takeaway

This AMC 12 problem only needs Grade 8 reciprocals and arithmetic sequences you already know! Take reciprocals: b_n = 1/a_n turns the messy rule into b_n = 2 b_n-1 - b_n-2, i.e., {b_n} is arithmetic. With b₁ = 1, b₂ = 7/3, the step is 4/3, so b_n = (4n - 1)/3. At n = 2019: b₂₀₁₉ = 8075/3, a₂₀₁₉ = 3/8075. Since 8075 = 5² · 17 · 19, the fraction is lowest, so p + q = 3 + 8075 = 8078, answer (E).