AMC 10 · 2017 · #13
Easy mode Grade 4A list of numbers starts with 0 and then 1. To make each new number, add the last two numbers and keep only the remainder after dividing by 3. So the list begins 0,1,1,2,0,2,….
Count the spots starting from 0: spot 0 holds 0, spot 1 holds 1, spot 2 holds 1, and so on. Add up the eight numbers in spots 2017,2018,2019,2020,2021,2022,2023, and 2024. What is the total?
Pick an answer.
AMC 10 2017 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 starts with $F_0=0$ and $F_1=1$, and after that each term $F_n$ is the remainder when $F_{n-1}+F_{n-2}$ is divided by $3$, so it begins $0,1,1,2,0,2,\ldots$. Find the value of the sum of the eight consecutive terms $F_{2017}+F_{2018}+F_{2019}+F_{2020}+F_{2021}+F_{2022}+F_{2023}+F_{2024}$.
Givens: $F_0=0$ and $F_1=1$; For $n\ge 2$, $F_n$ is the remainder when $F_{n-1}+F_{n-2}$ is divided by $3$; The sequence begins $0,1,1,2,0,2,\ldots$; Answer choices: (A) $6$, (B) $7$, (C) $8$, (D) $9$, (E) $10$
Unknowns: The sum of the eight terms from $F_{2017}$ through $F_{2024}$
Understand
Restated: A sequence starts with $F_0=0$ and $F_1=1$, and after that each term $F_n$ is the remainder when $F_{n-1}+F_{n-2}$ is divided by $3$, so it begins $0,1,1,2,0,2,\ldots$. Find the value of the sum of the eight consecutive terms $F_{2017}+F_{2018}+F_{2019}+F_{2020}+F_{2021}+F_{2022}+F_{2023}+F_{2024}$.
Givens: $F_0=0$ and $F_1=1$; For $n\ge 2$, $F_n$ is the remainder when $F_{n-1}+F_{n-2}$ is divided by $3$; The sequence begins $0,1,1,2,0,2,\ldots$; Answer choices: (A) $6$, (B) $7$, (C) $8$, (D) $9$, (E) $10$
Plan
Primary tool: #5 Look for a Pattern
Secondary: #9 Solve an Easier Related Problem
Because every term is just a remainder after dividing by $3$, each term can only be $0$, $1$, or $2$ — a tiny set — so the sequence is forced to repeat. Tool #5 (Look for a Pattern) tells us to write out terms until the starting pair comes back, which reveals the repeating block and its length. Once we know the period, Tool #9 (Solve an Easier Related Problem) lets us trade the impossible-looking terms $F_{2017}$ through $F_{2024}$ for one ordinary cycle: eight terms in a row on a length-$8$ loop is exactly one full lap, so the far-away sum equals the sum of one repeating block.
Execute — Answer: D
4.NBT.B.6 Step 1 List terms until the pair repeats
- Keep adding the previous two terms and writing only the remainder after dividing by $3$, so every term stays in $\{0,1,2\}$.
- From the given start: $F_0=0$, $F_1=1$, $F_2=1$, $F_3=2$, $F_4=0$ (since $2+1=3$ leaves remainder $0$), $F_5=2$, $F_6=2$, $F_7=1$.
- Then $F_8$ is the remainder of $F_7+F_6=1+2=3$, which is $0$, and $F_9$ is the remainder of $F_8+F_7=0+1=1$.
💡 Keeping only the remainder after dividing by $3$ traps every term inside the three values $0,1,2$, so the list cannot help but loop.
4.OA.C.5 Step 2 Find the repeating block of 8
- Look at $F_8=0$ and $F_9=1$: that is exactly the same pair that started the sequence, $F_0=0$ and $F_1=1$.
- Since each term depends only on the previous two, once a pair comes back the whole sequence repeats from there.
- So the block $0,1,1,2,0,2,2,1$ (terms $F_0$ through $F_7$) cycles forever, with period $8$.
💡 A two-back adding rule is steered entirely by the last two values, so the first repeated pair locks in an endless loop.
4.NBT.B.6 Step 3 Eight terms in a row make one full cycle
- The sum runs over the eight consecutive terms $F_{2017}$ through $F_{2024}$.
- Because the pattern repeats every $8$ terms, any $8$ terms in a row cover one complete cycle — one of each spot in the block — no matter where they start.
- To see exactly where they begin, divide: $2017 = 8\times 252 + 1$, so $F_{2017}$ sits at position $1$ of the block, and the next eight terms wrap once through positions $1,2,\ldots,7,0$.
💡 Eight stops on an eight-stop loop is exactly one trip around, wherever you board.
4.OA.A.3 Step 4 Add one full cycle
- So the sum equals the sum of one whole period: $0+1+1+2+0+2+2+1=9$.
- The answer is (D).
💡 Swap eight scattered far-away terms for the one tidy block they secretly copy, then just add.
4.NBT.B.6 Keep adding the previous two terms and writing only the remainder after dividing 4.OA.C.5 Look at $F_8=0$ and $F_9=1$: that is exactly the same pair that started the sequ 4.NBT.B.6 The sum runs over the eight consecutive terms $F_{2017}$ through $F_{2024}$. Bec 4.OA.A.3 So the sum equals the sum of one whole period: $0+1+1+2+0+2+2+1=9$. The answer i Review
Reasonableness: Every term is $0$, $1$, or $2$, so eight of them can total at most $8\times 2=16$ and at least $0$; the result $9$ sits comfortably inside the choices' range of $6$ to $10$. The period sum $0+1+1+2+0+2+2+1=9$ matches (D). Because the indices $2017$–$2024$ are eight in a row, the exact starting point doesn't matter — shifting all the way down to $F_0$–$F_7$ gives the same $9$.
Alternative: Tally the block instead of adding left to right: it holds two $0$'s, three $1$'s, and three $2$'s, so the sum is $1\times 3 + 2\times 3 = 3+6 = 9$, again (D).
CCSS standards used (min grade 4)
4.NBT.B.6Find whole-number quotients and remainders with up to four-digit dividends (Computing each term as the remainder of a sum divided by $3$, and dividing $2017$ by $8$ to locate its place in the cycle.)4.OA.C.5Generate a number or shape pattern following a given rule (Recognizing that the sequence repeats with period $8$ once the starting pair returns.)4.OA.A.3Solve multi-step word problems using four operations with whole numbers (Adding the eight terms of one full period to reach $9$.)
⭐ When a sequence loops, one full loop's worth of terms always adds to the same total — so find the loop length, then add up a single loop.
⭐ When a sequence loops, one full loop's worth of terms always adds to the same total — so find the loop length, then add up a single loop.
More like this
Same archetype — closest grade level first.