AMC 10 · 2021 · #8
Easy mode Grade 4A list of numbers starts 0, 0, 1. Call those spots position 0, position 1, and position 2.
Every number after that follows one rule: it is the number just before it plus the number three positions earlier. So position 3 holds 1+0=1.
The numbers at positions 2021, 2022, and 2023 are far too big to write out, but we only need to know whether each one is even or odd. Writing E for even and O for odd, which choice lists those three answers in that order?
Pick an answer.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: A sequence starts $D_0 = 0$, $D_1 = 0$, $D_2 = 1$, and from then on every term is the previous term plus the term three places back: $D_n = D_{n-1} + D_{n-3}$. We must say whether each of $D_{2021}$, $D_{2022}$, $D_{2023}$ is even ($E$) or odd ($O$), in that order.
Givens: Starting values $D_0 = 0$, $D_1 = 0$, $D_2 = 1$; Recursion $D_n = D_{n-1} + D_{n-3}$ for every $n \ge 3$; Only the parity (even or odd) of each term is asked for, not its value; Choices: (A) $(O,E,O)$, (B) $(E,E,O)$, (C) $(E,O,E)$, (D) $(O,O,E)$, (E) $(O,O,O)$
Unknowns: The parity of $D_{2021}$; The parity of $D_{2022}$; The parity of $D_{2023}$
Understand
Restated: A sequence starts $D_0 = 0$, $D_1 = 0$, $D_2 = 1$, and from then on every term is the previous term plus the term three places back: $D_n = D_{n-1} + D_{n-3}$. We must say whether each of $D_{2021}$, $D_{2022}$, $D_{2023}$ is even ($E$) or odd ($O$), in that order.
Givens: Starting values $D_0 = 0$, $D_1 = 0$, $D_2 = 1$; Recursion $D_n = D_{n-1} + D_{n-3}$ for every $n \ge 3$; Only the parity (even or odd) of each term is asked for, not its value; Choices: (A) $(O,E,O)$, (B) $(E,E,O)$, (C) $(E,O,E)$, (D) $(O,O,E)$, (E) $(O,O,O)$
Plan
Primary tool: #5 Look for a Pattern
Secondary: #9 Solve an Easier Related Problem, #2 Make a Systematic List, #3 Eliminate Possibilities
Tool #9 (Easier Related Problem) makes the whole thing tractable: throw away the actual sizes and keep only even/odd, because parity of a sum is decided by parity of the parts. Tool #2 (Systematic List) builds the even/odd table from $n = 0$ upward without skipping a term. Tool #5 (Look for a Pattern) is the payoff — the parity list repeats, and once the pattern's length is known, index $2021$ collapses to a tiny index. Tool #3 (Eliminate Possibilities) confirms the resulting triple against the five listed options.
Execute — Answer: C
2.OA.C.3 Step 1 Track parity, not size
- $D_{2021}$ is far too big to write down, but the question only asks even or odd.
- The parity of a sum is completely determined by the parities of the two numbers added: even plus even is even, even plus odd is odd, odd plus odd is even.
- So the rule $D_n = D_{n-1} + D_{n-3}$ can be run using only the letters $E$ and $O$, never the numbers themselves.
💡 Even and odd behave like a two-button calculator, so a hopeless computation shrinks to a game with two symbols.
4.OA.C.5 Step 2 Build the first terms
Run the recursion forward from the three starting values, one term at a time: $D_3 = D_2 + D_0 = 1$, $D_4 = D_3 + D_1 = 1$, $D_5 = D_4 + D_2 = 2$, $D_6 = D_5 + D_3 = 3$, $D_7 = D_6 + D_4 = 4$, $D_8 = D_7 + D_5 = 6$, $D_9 = D_8 + D_6 = 9$, $D_{10} = D_9 + D_7 = 13$.
💡 Generating a handful of terms from the stated rule is the cheapest way to see what the sequence actually does.
4.OA.C.5 Step 3 Write the parity row
- Replace each term by $E$ or $O$.
- From $n = 0$: $0,0,1,1,1,2,3,4,6,9,13$ becomes $E,E,O,O,O,E,O,E,E,O,O$.
- Reading indices $0$ through $6$ gives the block $E,E,O,O,O,E,O$.
💡 Stripping the table down to two symbols makes a repeat visible that the raw numbers hide.
3.OA.D.9 Step 4 Spot the repeat: period 7
- Compare the parities at $n = 7,8,9$ with those at $n = 0,1,2$: both triples are $(E,E,O)$.
- Each new term is built from the three most recent terms, so once a triple of consecutive parities comes back, every parity after it must repeat too.
- The parity sequence therefore cycles with period $7$: $D_n$ and $D_{n+7}$ always have the same parity.
💡 A rule that looks back only three steps has no memory beyond that, so a repeated three-term state forces the whole future to repeat.
4.NBT.B.6 Step 5 Reduce 2021, 2022, 2023 mod 7
- With period $7$, only the remainder of the index on division by $7$ matters.
- Since $7 \times 288 = 2016$, we get $2021 = 7 \times 288 + 5$, so $2021$ leaves remainder $5$.
- The next two indices leave remainders $6$ and $0$.
💡 Periodicity turns a four-digit index into its remainder, which is all the cycle can see.
4.OA.C.5 Step 6 Read the answer off the table
- Look up positions $5$, $6$, $0$ in the parity block $E,E,O,O,O,E,O$: position $5$ is $E$, position $6$ is $O$, position $0$ is $E$.
- So $(D_{2021}, D_{2022}, D_{2023})$ has parities $(E, O, E)$, which is choice (C).
💡 Once the index is reduced, the answer is a plain table lookup.
2.OA.C.3 $D_{2021}$ is far too big to write down, but the question only asks even or odd. 4.OA.C.5 Run the recursion forward from the three starting values, one term at a time: $D 4.OA.C.5 Replace each term by $E$ or $O$. From $n = 0$: $0,0,1,1,1,2,3,4,6,9,13$ becomes 3.OA.D.9 Compare the parities at $n = 7,8,9$ with those at $n = 0,1,2$: both triples are 4.NBT.B.6 With period $7$, only the remainder of the index on division by $7$ matters. Sin 4.OA.C.5 Look up positions $5$, $6$, $0$ in the parity block $E,E,O,O,O,E,O$: position $5 Review
Reasonableness: The period-7 claim is checkable beyond the one match used to prove it: parities at $n = 10,11,12,13$ are $O,O,E,O$ (from $13, 19, 28, 41$), exactly matching $n = 3,4,5,6$, which were $O,O,E,O$. Extending the table to $n = 20$ keeps the block $E,E,O,O,O,E,O$ repeating with no drift. The three answer indices land on three different positions of the cycle, and the resulting triple $(E,O,E)$ appears exactly once among the five options, so no other choice is compatible. As a sanity note on plausibility, $D_{2023} = D_{2022} + D_{2020}$, and an even result requires its two addends to share a parity — consistent with the cycle, which puts $D_{2022}$ at position $6$ ($O$) and $D_{2020}$ at position $4$ ($O$).
Alternative: Tool #3 (Eliminate Possibilities) can prune before any long table is built. From the recursion, $D_{2023} = D_{2022} + D_{2020}$ and $D_{2022} = D_{2021} + D_{2019}$, so relations among consecutive terms constrain which triples are even possible. Alternatively one can work strictly in modular arithmetic, writing the rule as $d_n \equiv d_{n-1} + d_{n-3} \pmod 2$ and listing the $8$ possible states $(d_{n}, d_{n+1}, d_{n+2})$; since only finitely many states exist, the sequence must cycle, and following the states from $(0,0,1)$ returns to $(0,0,1)$ after exactly $7$ moves. That state-machine view proves the period must exist before any terms are computed.
CCSS standards used (min grade 4)
2.OA.C.3Determine whether a group of objects has an odd or even number (Using the even/odd addition rules $E+E=E$, $E+O=O$, $O+O=E$ so the recursion can be run on parities alone.)4.OA.C.5Generate a number or shape pattern following a given rule (Generating $D_0$ through $D_{10}$ from $D_n = D_{n-1} + D_{n-3}$ and reading the matching parity off the table.)3.OA.D.9Identify arithmetic patterns and explain using properties of operations (Recognizing that the repeated triple $(E,E,O)$ at $n = 7,8,9$ forces the parity sequence to cycle with period $7$.)4.NBT.B.6Find whole-number quotients and remainders with up to four-digit dividends (Dividing $2021$, $2022$, $2023$ by $7$ to get remainders $5$, $6$, $0$.)
⭐ This AMC 12 problem needs no algebra beyond Grade 4: keep only even/odd instead of the huge numbers, list the first few terms, notice the pattern repeats every 7 steps, then divide 2021 by 7 and look up the remainder.
⭐ This AMC 12 problem needs no algebra beyond Grade 4: keep only even/odd instead of the huge numbers, list the first few terms, notice the pattern repeats every 7 steps, then divide 2021 by 7 and look up the remainder.
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