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2017 AMC 10A Problem 13

2017 AMC 10A Problem 13AMC 10 Step by Step

Name: Date: September 13, 2026

Modular Arithmetic · difficulty 2 of 5 · about 3 min · position P11-15

Define a sequence recursively by $F_{0}=0,~F_{1}=1,$ and $F_{n}=$ the remainder when $F_{n-1}+F_{n-2}$ is divided by $3,$ for all $n\geq 2.$ Thus the sequence starts $0,1,1,2,0,2,\ldots$ What is $F_{2017}+F_{2018}+F_{2019}+F_{2020}+F_{2021}+F_{2022}+F_{2023}+F_{2024}?$

A)
$6$
B)
$7$
C)
$8$
D)
$9$
E)
$10$

Answer

Key insight

Fibonacci mod 3 repeats with period 8, so any eight consecutive terms are a full cycle and sum to 0+1+1+2+0+2+2+1 = 9.

Solution

Continue the sequence, reducing each sum mod $3$ :
$0,\ 1,\ 1,\ 2,\ 0,\ 2,\ 2,\ 1,\ 0,\ 1,\ 1,\ 2,\ \ldots$
After eight terms the pair $(0, 1)$ reappears at $F_8, F_9$ . Since each term depends only on the previous two, the sequence repeats from there with period $8$ .

The requested sum has exactly eight consecutive terms, so it covers one full period, in some rotation. Its value is the sum of one period:
$0 + 1 + 1 + 2 + 0 + 2 + 2 + 1 = 9.$

The answer is $\boxed{\textbf{(D)}\ 9}$ .

Why this works

A recursion taken mod $m$ has only finitely many possible states (here $3 \cdot 3 = 9$ pairs), so it must cycle; you find the period by watching for the initial pair to return. Once the period is known, a sum over a block whose length is a multiple of the period does not depend on where the block starts, which is why the specific index $2017$ never matters.

The trap

Stopping the pattern search too early (thinking the period is 4 after seeing 0,1,1,2,0) or computing 2017 mod 8 when it is not even needed.

Common mistakes

Stopping the pattern search too early (thinking the period is 4 after seeing 0,1,1,2,0) or computing 2017 mod 8 when it is not even needed.
Summing seven or nine terms because of an off-by-one count of the indices $2017$ through $2024$ .

Techniques

Compute small cases, spot the pattern, generalize