AMC 10 · 2019 · #15
Grade 8 number-theoryProblem
A sequence of numbers is defined recursively by a1=1, a2=73, and
an=2an−2−an−1an−2⋅an−1for all n≥3 Then a2019 can be written as qp, where p and q are relatively prime positive integers. What is p+q ?
Pick an answer.
AMC 10 2019 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: A sequence is given by $a_1 = 1$, $a_2 = \frac{3}{7}$, and the recursion $a_n = \dfrac{a_{n-2} \cdot a_{n-1}}{2a_{n-2} - a_{n-1}}$ for $n \ge 3$. The value $a_{2019}$ equals $\frac{p}{q}$ in lowest terms with $p, q$ positive integers. Find $p + q$.
Givens: $a_1 = 1,\;\; a_2 = \tfrac{3}{7}$; $a_n = \dfrac{a_{n-2} \cdot a_{n-1}}{2 a_{n-2} - a_{n-1}}$ for $n \ge 3$; $a_{2019} = \tfrac{p}{q}$ in lowest terms; Answer choices: $2020, 4039, 6057, 6061, 8078$
Unknowns: $p + q$
Understand
Restated: A sequence is given by $a_1 = 1$, $a_2 = \frac{3}{7}$, and the recursion $a_n = \dfrac{a_{n-2} \cdot a_{n-1}}{2a_{n-2} - a_{n-1}}$ for $n \ge 3$. The value $a_{2019}$ equals $\frac{p}{q}$ in lowest terms with $p, q$ positive integers. Find $p + q$.
Givens: $a_1 = 1,\;\; a_2 = \tfrac{3}{7}$; $a_n = \dfrac{a_{n-2} \cdot a_{n-1}}{2 a_{n-2} - a_{n-1}}$ for $n \ge 3$; $a_{2019} = \tfrac{p}{q}$ in lowest terms; Answer choices: $2020, 4039, 6057, 6061, 8078$
Plan
Primary tool: #15 Organize Information in More Ways
Secondary: #5 Look for a Pattern, #9 Solve an Easier Related Problem, #3 Eliminate Possibilities
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 $a_3, a_4$ by hand so the arithmetic pattern in $b$ is visible. Then $b_{2019}$ follows from an arithmetic-sequence formula, and the answer is $p + q$. Tool #3 matches the result to the five choices.
Execute — Answer: E
8.EE.A.1 Step 1 - Reorganize: divide both sides of the recursion by $a_{n-2} a_{n-1} a_n$ to make reciprocals appear.
- From $a_n = \dfrac{a_{n-2} a_{n-1}}{2 a_{n-2} - a_{n-1}}$, take reciprocals: $\dfrac{1}{a_n} = \dfrac{2 a_{n-2} - a_{n-1}}{a_{n-2} a_{n-1}} = \dfrac{2}{a_{n-1}} - \dfrac{1}{a_{n-2}}$.
- Let $b_n = \dfrac{1}{a_n}$.
- The recursion becomes $b_n = 2 b_{n-1} - b_{n-2}$.
💡 Grade 8 exponents/algebra: flipping a multiplicative rule into a reciprocal rule trades a messy product for a clean linear combination.
4.OA.C.5 Step 2 - Rewrite $b_n = 2 b_{n-1} - b_{n-2}$ as $b_n - b_{n-1} = b_{n-1} - b_{n-2}$.
- The consecutive differences are all equal, so $\{b_n\}$ is an arithmetic sequence.
💡 Grade 8 patterns: equal differences mean each new term is found by adding the same step.
4.OA.C.5 Step 3 - Compute $b_1$ and $b_2$ to find the common difference.
- $b_1 = 1/a_1 = 1$ and $b_2 = 1/a_2 = 7/3$.
- Common difference $d = b_2 - b_1 = \tfrac{7}{3} - 1 = \tfrac{4}{3}$.
- So $b_n = 1 + (n-1) \cdot \tfrac{4}{3} = \tfrac{3 + 4(n-1)}{3} = \tfrac{4n - 1}{3}$.
💡 Grade 8 arithmetic sequence: explicit formula is starting value plus $(n-1)$ times the common difference.
5.NF.B.7 Step 4 - Sanity-check with the easier $n = 3$ case.
- $b_3 = \tfrac{4 \cdot 3 - 1}{3} = \tfrac{11}{3}$, so $a_3 = \tfrac{3}{11}$.
- From the original recursion: $a_3 = \dfrac{a_1 a_2}{2 a_1 - a_2} = \dfrac{1 \cdot \tfrac{3}{7}}{2 - \tfrac{3}{7}} = \dfrac{\tfrac{3}{7}}{\tfrac{11}{7}} = \dfrac{3}{11}$.
- Matches.
💡 Grade 5 fraction division: a small test confirms the explicit formula.
6.EE.A.2 Step 5 - Set $n = 2019$.
- $b_{2019} = \dfrac{4 \cdot 2019 - 1}{3} = \dfrac{8076 - 1}{3} = \dfrac{8075}{3}$.
- Therefore $a_{2019} = \dfrac{1}{b_{2019}} = \dfrac{3}{8075}$.
💡 Grade 6 expressions: plug in $n = 2019$ into the formula.
6.NS.B.4 Step 6 - Check that $\tfrac{3}{8075}$ is already in lowest terms.
- Factor: $8075 = 25 \cdot 323 = 25 \cdot 17 \cdot 19 = 5^2 \cdot 17 \cdot 19$.
- None of those factors is $3$, so $\gcd(3, 8075) = 1$.
- Hence $p = 3,\; q = 8075$.
💡 Grade 6 GCF: prime factorize and check $3$ doesn't appear.
4.NBT.B.4 Step 7 Compute $p + q = 3 + 8075 = 8078$.
💡 Grade 4 multi-digit addition: $3 + 8075 = 8078$.
4.NBT.A.2 Step 8 Match $8078$ to the choices: $(E)$.
💡 Grade 4: pick the matching value from the list.
8.EE.A.1 Reorganize: divide both sides of the recursion by $a_{n-2} a_{n-1} a_n$ to make 4.OA.C.5 Rewrite $b_n = 2 b_{n-1} - b_{n-2}$ as $b_n - b_{n-1} = b_{n-1} - b_{n-2}$. The 4.OA.C.5 Compute $b_1$ and $b_2$ to find the common difference. $b_1 = 1/a_1 = 1$ and $b_ 5.NF.B.7 Sanity-check with the easier $n = 3$ case. $b_3 = \tfrac{4 \cdot 3 - 1}{3} = \tf 6.EE.A.2 Set $n = 2019$. $b_{2019} = \dfrac{4 \cdot 2019 - 1}{3} = \dfrac{8076 - 1}{3} = 6.NS.B.4 Check that $\tfrac{3}{8075}$ is already in lowest terms. Factor: $8075 = 25 \cdo 4.NBT.B.4 Compute $p + q = 3 + 8075 = 8078$. 4.NBT.A.2 Match $8078$ to the choices: $(E)$. Review
Reasonableness: The answer choices give a sharp clue: $8078 = 2 \cdot 4039 \approx 4 \cdot 2019$. That matches $q = 8075$, since $b_{2019} = \tfrac{4 \cdot 2019 - 1}{3}$ has numerator $8075 \approx 4 \cdot 2019$. Also $a_n \to 0$ as $n$ grows (since $b_n \to \infty$), which is consistent with $\tfrac{3}{8075}$ being tiny.
Alternative: Tool #6 (Guess and Check) by computing $a_3, a_4, a_5$ directly from the recursion. $a_3 = 3/11$, $a_4 = 3/15 = 1/5$, $a_5 = 3/19$, $a_6 = 3/23$. Pattern: $a_n = \dfrac{3}{4n - 1}$ for $n \ge 2$ (with $a_1 = 1$ matching at $n = 1$ as $3/3$). At $n = 2019$: $a_{2019} = 3/8075$. Same answer.
CCSS standards used (min grade 8)
4.NBT.A.2Read and write multi-digit whole numbers and compare using symbols (Matching the final value $8078$ to the answer choices.)4.NBT.B.4Fluently add and subtract multi-digit whole numbers (Computing $p + q = 3 + 8075 = 8078$.)4.OA.C.5Generate a number or shape pattern following a given rule (Recognizing that $b_n - b_{n-1} = $ constant gives the arithmetic-sequence formula $b_n = (4n - 1)/3$.)5.NF.B.7Apply and extend understanding of division to divide unit fractions by whole numbers (Sanity-checking $a_3 = (3/7)/(11/7) = 3/11$.)6.EE.A.2Write, read, and evaluate expressions in which letters stand for numbers (Evaluating $b_n = (4n - 1)/3$ at $n = 2019$.)6.NS.B.4Find greatest common factor and least common multiple of two numbers (Confirming $\gcd(3, 8075) = 1$ via the factorization $8075 = 5^2 \cdot 17 \cdot 19$.)8.EE.A.1Know and apply the properties of integer exponents (Reorganizing the multiplicative recursion into a reciprocal-form linear recursion $b_n = 2 b_{n-1} - b_{n-2}$.)
⭐ This AMC 10 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 $b_1 = 1, b_2 = 7/3$, the step is $4/3$, so $b_n = (4n - 1)/3$. At $n = 2019$: $b_{2019} = 8075/3$, $a_{2019} = 3/8075$. Since $8075 = 5^2 \cdot 17 \cdot 19$, the fraction is lowest, so $p + q = 3 + 8075 = \mathbf{8078}$, answer $(E)$.
⭐ This AMC 10 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 $b_1 = 1, b_2 = 7/3$, the step is $4/3$, so $b_n = (4n - 1)/3$. At $n = 2019$: $b_{2019} = 8075/3$, $a_{2019} = 3/8075$. Since $8075 = 5^2 \cdot 17 \cdot 19$, the fraction is lowest, so $p + q = 3 + 8075 = \mathbf{8078}$, answer $(E)$.
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