AMC 10 · 2019 · #15

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 is given by a₁ = 1, a₂ =

\frac{3}{7}

, and the recursion a_n =

\frac{a_n-2 · a_n-1}{2a_n-2 - a_n-1}

for n ≥ 3. The value a₂019 equals

\frac{p}{q}

in lowest terms with p, q positive integers. Find p + q.

Pick an answer.

(A)

2020

(B)

4039

(C)

6057

(D)

6061

(E)

8078

AMC 10 2019 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Organize Information in More Ways

Tool #15 (Reorganize): take reciprocals — set b_n = $\frac{1}{a_n}$ . 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: with b_n = $\frac{1}{a_n}$ the messy product becomes the clean linear rule b_n = 2 b_n-1 - b_n-2.

b_n =

\frac{1}{a_n}

→ b_n = 2 b_n-1 - b_n-2

2STEP 2

Rewrite it as b_n - b_n-1 = b_n-1 - b_n-2: the differences are all equal, so {b_n} is arithmetic.

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

3STEP 3

From b₁ = 1 and b₂ = $\frac{7}{3}$ the common difference is $\frac{4}{3}$ , giving b_n = $\frac{4n - 1}{3}$ .

b₁ = 1, b₂ =

\frac{7}{3}

, d =

\frac{4}{3}

→ b_n =

\frac{4n - 1}{3}

4STEP 4

Sanity-check at n = 3: the formula gives a₃ = $\frac{3}{11}$ , and the original recursion gives the same value.

a₃ = (

\frac{3}{7}

)/(

\frac{11}{7}

) =

\frac{3}{11}

\frac{1}{b₃}

✓

5STEP 5

At n = 2019: b₂019 = $\frac{4 · 2019 - 1}{3}$ = $\frac{8075}{3}$ , so a₂019 = $\frac{3}{8075}$ .

b₂019 =

\frac{8075}{3}

, a₂019 =

\frac{3}{8075}

6STEP 6

Since 8075 = 5² · 17 · 19 has no factor of 3, the fraction is already lowest: p = 3, q = 8075.

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

7STEP 7

Add: p + q = 3 + 8075 = 8078.

p + q = 3 + 8075 = 8078

8STEP 8

8078 matches choice (E).

8078 → (E)

Answer

8078

The answer choices give a sharp clue: 8078 = 2 · 4039 ≈ 4 · 2019. That matches q = 8075, since b₂019 =

\frac{4 · 2019 - 1}{3}

has numerator 8075 ≈ 4 · 2019. Also a_n → 0 as n grows (since b_n → ∞), which is consistent with

\frac{3}{8075}

being tiny.

💡Key takeaway

This AMC 10 problem only needs Grade 8 reciprocals and arithmetic sequences you already know! Take reciprocals: b_n = $\frac{1}{a_n}$ turns the messy rule into b_n = 2 b_n-1 - b_n-2, i.e., {b_n} is arithmetic. With b₁ = 1, b₂ = $\frac{7}{3}$ , the step is $\frac{4}{3}$ , so b_n = $\frac{4n - 1}{3}$ . At n = 2019: b₂019 = $\frac{8075}{3}$ , a₂019 = $\frac{3}{8075}$ . Since 8075 = 5² · 17 · 19, the fraction is lowest, so p + q = 3 + 8075 = 8078, answer (E).

More like this