AMC 10 · 2018 · #20

Grade 4 arithmetic
Archetype Lattice / Grid Pattern from Small Cases └ Sub-archetype Periodic Placement Formula
recursive-sequencepattern-recognition pattern-recognitioneasier-related-problem ↑ Prerequisites: recursive-sequence
📏 Medium solution 💡 2 insights
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Problem
A list of numbers is built by a rule: the first two are both 1, and from the third on, each number equals the one before it, minus the one before that, plus its own position number. Find the 2018th number on the list.

Pick an answer.

(A)
2016
(B)
2017
(C)
2018
(D)
2019
(E)
2020

AMC 10 2018 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Look for a Pattern

Tool #2 (Make a Systematic List): the rule is easy to apply, so writing out the first dozen terms costs almost nothing and gives real numbers to stare at. Tool #5 (Look for a Pattern): once the early terms are listed, the key is to compare terms six apart and notice they climb by a fixed amount — that single observation replaces 2018 calculations. Tool #9 (Solve an Easier Related Problem): with the six-step jump in hand, finding f(2018) collapses to finding a tiny term near the start plus a count of how many jumps fit in 2018.

1STEP 1

List the first several terms

Apply the rule from f(1)=f(2)=1: the early terms come out 1, 1, 3, 6, 8, 8, 7, 7, 9, 12, 14, 14, 13.

f(1{…}13) = 1, 1, 3, 6, 8, 8, 7, 7, 9, 12, 14, 14, 13
2STEP 2

Compare terms six apart

Pair each term with the one six later — f(1),f(7); f(2),f(8); … — every pair differs by 6, so six positions ahead adds 6 to the value.

f(n+6) = f(n) + 6
3STEP 3

Find how many six-jumps reach 2018

Dividing 2018 by 6 gives 2018 = 6 × 336 + 2, so f(2018) sits 336 six-jumps above the small term f(2).

2018 = 6 × 336 + 2 → f(2018) = f(2) + 6 × 336
4STEP 4

Add it up

With f(2) = 1 and 6 × 336 = 2016 added, f(2018) = 1 + 2016 = 2017, matching choice (B).

f(2018) = 1 + 6 × 336 = 1 + 2016 = 2017 → (B)
Answer
2017
The answer 2017 sits just below the position number 2018, which feels right: the listed terms stay close to their position number (for example f(7) = 7, f(13) = 13), so f(2018) should be near 2018. The four trap choices 2016, 2018, 2019, 2020 are all one or two off — exactly the kind of slip you get from miscounting the jumps or starting from the wrong small term. A quick check pins it down: 2016 = 6 × 336 comes from the 336 jumps, and the leftover +1 comes from f(2) = 1, giving 2017 with no rounding room.
💡Key takeaway

When a sequence repeats its shape every six steps, jump six at a time: count the jumps, add six for each, and start from a small known term.

  • List the first several terms
  • Compare terms six apart
  • Find how many six-jumps reach 2018
  • Add it up

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