AMC 10 · 2023 · #6
Grade 4 arithmeticProblem
Let L1=1,L2=3, and Ln+2=Ln+1+Ln for n≥1. How many terms in the sequence L1,L2,L3,...,L2023 are even?
Pick an answer.
AMC 10 2023 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 Lucas-style sequence is defined by $L_1 = 1$, $L_2 = 3$, and $L_{n+2} = L_{n+1} + L_n$. Count how many of $L_1, L_2, \ldots, L_{2023}$ are even.
Givens: $L_1 = 1$ (odd); $L_2 = 3$ (odd); Recurrence: each later term is the sum of the two before it; Range to scan: $L_1$ through $L_{2023}$ — that is $2023$ terms; Answer choices: (A) $673$, (B) $1011$, (C) $675$, (D) $1010$, (E) $674$
Unknowns: How many of those $2023$ terms are even
Understand
Restated: A Lucas-style sequence is defined by $L_1 = 1$, $L_2 = 3$, and $L_{n+2} = L_{n+1} + L_n$. Count how many of $L_1, L_2, \ldots, L_{2023}$ are even.
Givens: $L_1 = 1$ (odd); $L_2 = 3$ (odd); Recurrence: each later term is the sum of the two before it; Range to scan: $L_1$ through $L_{2023}$ — that is $2023$ terms; Answer choices: (A) $673$, (B) $1011$, (C) $675$, (D) $1010$, (E) $674$
Plan
Primary tool: #5 Look for a Pattern
Secondary: #9 Solve an Easier Related Problem, #2 Make a Systematic List
Counting parities for $2023$ terms is hopeless by brute force, so reach for Tool #9 (Easier Problem): compute the first few parities by hand. As soon as the odd/even tags repeat the starting pair $(\text{odd}, \text{odd})$, Tool #5 (Pattern) takes over — the cycle must continue forever. Tool #2 (Systematic List) keeps the parity record tidy. Once the period is known, the original $2023$ shrinks to a one-line division problem: how many cycle-ending slots fit in $2023$. Algebra (Tool #13) is unnecessary; the parity rules already do all the work.
Execute — Answer: E
2.OA.C.3 Step 1 - List the parity of the first several terms using the rule Odd+Odd=Even, Odd+Even=Odd, Even+Odd=Odd.
- Start with the two givens $L_1, L_2$ both odd, then apply the rule term by term.
💡 Working with a much smaller sample of the sequence makes the structure visible — that is Grade 2 odd/even labeling.
4.OA.C.5 Step 2 - The parity row reads O, O, E, O, O, E, O, O, ...
- — the pattern (O, O, E) appears to repeat with period $3$.
- Because each term's parity depends only on the two before it, the moment a parity pair $(L_n, L_{n+1})$ repeats an earlier pair the whole tail repeats.
- Pair $(L_4, L_5) = (O, O)$ matches $(L_1, L_2) = (O, O)$, so the period is exactly $3$.
💡 Spotting a repeating block of three and arguing it must continue is exactly Grade 4 pattern-rule reasoning.
4.OA.B.4 Step 3 - Inside each block the even term is in the third slot.
- So $L_n$ is even exactly when $n$ is a multiple of $3$.
- The question now reduces to: how many multiples of $3$ are in $\{1, 2, \ldots, 2023\}$?
💡 Recognizing the even slots are exactly the multiples of $3$ is Grade 4 multiples thinking.
4.NBT.B.6 Step 4 - Count multiples of $3$ up to $2023$ by dividing.
- The largest multiple of $3$ that fits is $3 \times 674 = 2022$, and $3 \times 675 = 2025$ is already too big.
- So there are $674$ multiples of $3$ in $\{1, \ldots, 2023\}$.
💡 One division with remainder gives the count — Grade 4 multi-digit division.
2.OA.C.3 List the parity of the first several terms using the rule Odd+Odd=Even, Odd+Even 4.OA.C.5 The parity row reads O, O, E, O, O, E, O, O, ... — the pattern (O, O, E) appears 4.OA.B.4 Inside each block the even term is in the third slot. So $L_n$ is even exactly w 4.NBT.B.6 Count multiples of $3$ up to $2023$ by dividing. The largest multiple of $3$ tha Review
Reasonableness: Quick sanity: roughly $\tfrac{1}{3}$ of the $2023$ terms should be even, and $\tfrac{2023}{3} \approx 674.3$ — so an answer near $674$ is expected. (E) $674$ matches. Choice (B) $1011$ is what you would get if you mistakenly counted odd-indexed terms; (C) $675$ is the off-by-one trap of rounding up; (D) $1010$ would be $\tfrac{2020}{2}$ — a halving error rather than a thirding. The parity-block argument also passes the endpoint test: $L_3 = 4$ (even ✓), $L_6 = 18$ (even ✓), $L_9 = L_8 + L_7 = 47 + 29 = 76$ (even ✓).
Alternative: Tool #13 (Convert to Algebra) via modular arithmetic: work the recurrence modulo $2$. Starting from $L_1 \equiv 1, L_2 \equiv 1 \pmod{2}$, get $L_3 \equiv 0$, $L_4 \equiv 1$, $L_5 \equiv 1$, $L_6 \equiv 0$. The pair $(1,1)$ resets, proving period $3$. Same conclusion, but with formal $\bmod 2$ notation — heavier machinery than the parity-block picture for the same answer.
CCSS standards used (min grade 4)
2.OA.C.3Determine whether a group of objects has an odd or even number (Labeling each computed Lucas-style term as odd or even — the only feature of the term that matters for this problem.)4.OA.C.5Generate a number or shape pattern following a given rule (Reading the repeating block (O, O, E) out of the parity row and arguing the rule forces the same block to recur forever.)4.OA.B.4Find all factor pairs and recognize multiples; determine prime or composite (Identifying that the even-term positions $3, 6, 9, \ldots$ are exactly the multiples of $3$.)4.NBT.B.6Find whole-number quotients and remainders with up to four-digit dividends (Dividing $2023 \div 3 = 674$ remainder $1$ to count multiples of $3$ at or below $2023$.)
⭐ This AMC 10 problem only needs Grade 4 pattern and division skills you already know — write down odd or even for the first few terms, spot the (odd, odd, even) block of three, then ask how many of those blocks fit inside $2023$.
⭐ This AMC 10 problem only needs Grade 4 pattern and division skills you already know — write down odd or even for the first few terms, spot the (odd, odd, even) block of three, then ask how many of those blocks fit inside $2023$.
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