AMC 10 · 2018 · #20
Easy mode Grade 4You build a long list of numbers, one at a time. The 1st number is 1 and the 2nd number is 1. From the 3rd number on, you find each number with this rule: take the number right before it, subtract the number two spots before it, then add the spot number itself.
So the number in spot n is f(n)=f(n−1)−f(n−2)+n.
What is the number in spot 2018?
Pick an answer.
AMC 10 2018 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 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 $2018$th number on the list.
Givens: $f(1) = 1$ and $f(2) = 1$; For every $n \ge 3$, $f(n) = f(n-1) - f(n-2) + n$; Answer choices: (A) $2016$, (B) $2017$, (C) $2018$, (D) $2019$, (E) $2020$
Unknowns: The value of $f(2018)$
Understand
Restated: 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 $2018$th number on the list.
Givens: $f(1) = 1$ and $f(2) = 1$; For every $n \ge 3$, $f(n) = f(n-1) - f(n-2) + n$; Answer choices: (A) $2016$, (B) $2017$, (C) $2018$, (D) $2019$, (E) $2020$
Plan
Primary tool: #5 Look for a Pattern
Secondary: #2 Make a Systematic List, #9 Solve an Easier Related Problem
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 $2018$.
Execute — Answer: B
4.OA.C.5 Step 1 List the first several terms
- Start from the two given terms and apply the rule one step at a time.
- $f(3) = f(2) - f(1) + 3 = 1 - 1 + 3 = 3$.
- $f(4) = 3 - 1 + 4 = 6$.
- $f(5) = 6 - 3 + 5 = 8$.
- $f(6) = 8 - 6 + 6 = 8$.
- $f(7) = 8 - 8 + 7 = 7$.
- $f(8) = 7 - 8 + 8 = 7$.
- Keep going: $f(9) = 9$, $f(10) = 12$, $f(11) = 14$, $f(12) = 14$, $f(13) = 13$.
💡 The rule only needs the two previous numbers, so each new term is one quick add-and-subtract.
3.OA.D.9 Step 2 Compare terms six apart
- Line up each term with the one six positions later.
- $f(1) = 1$ and $f(7) = 7$.
- $f(2) = 1$ and $f(8) = 7$.
- $f(3) = 3$ and $f(9) = 9$.
- $f(4) = 6$ and $f(10) = 12$.
- $f(5) = 8$ and $f(11) = 14$.
- $f(6) = 8$ and $f(12) = 14$.
- Every pair differs by exactly $6$.
- So jumping ahead six positions always adds $6$ to the value.
💡 The position number $n$ grows by $6$ over six steps, and the rest of the rule keeps repeating, so the value just rides up by $6$.
4.NBT.B.6 Step 3 Find how many six-jumps reach 2018
- To reach position $2018$ from a small starting position, keep adding $6$.
- Divide $2018$ by $6$: $2018 = 6 \times 336 + 2$.
- The remainder $2$ means $2018$ lands on the same spot in the cycle as position $2$, after $336$ jumps of six.
- So $f(2018) = f(2) + 6 \times 336$.
💡 Each jump of six adds six to both the position and the value, so counting the jumps counts the total added.
4.OA.A.3 Step 4 Add it up
- Plug in the small starting value $f(2) = 1$ and the $336$ jumps.
- Each jump adds $6$, contributing $6 \times 336 = 2016$.
- Then $f(2018) = 1 + 2016 = 2017$.
- That matches choice (B).
💡 Start from a known small term and add six for every full jump to the target position.
4.OA.C.5 Start from the two given terms and apply the rule one step at a time. $f(3) = f( 3.OA.D.9 Line up each term with the one six positions later. $f(1) = 1$ and $f(7) = 7$. $ 4.NBT.B.6 To reach position $2018$ from a small starting position, keep adding $6$. Divide 4.OA.A.3 Plug in the small starting value $f(2) = 1$ and the $336$ jumps. Each jump adds Review
Reasonableness: 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: $2016 = 6 \times 336$ comes from the $336$ jumps, and the leftover $+1$ comes from $f(2) = 1$, giving $2017$ with no rounding room.
Alternative: Subtract off the position to simplify. Let $g(n) = f(n) - n$. The rule becomes $g(n) = g(n-1) - g(n-2)$, a pattern with no $n$ in it, and such patterns repeat every six terms. Since $g(2) = f(2) - 2 = -1$ and $2018$ shares position $2$'s spot in the six-cycle, $g(2018) = g(2) = -1$, so $f(2018) = 2018 + g(2018) = 2018 - 1 = 2017$.
CCSS standards used (min grade 4)
4.OA.C.5Generate a number or shape pattern following a given rule (Building the first thirteen terms by applying the recursive rule one step at a time.)3.OA.D.9Identify arithmetic patterns and explain using properties of operations (Spotting that terms six positions apart always differ by $6$, i.e. $f(n+6) = f(n) + 6$.)4.NBT.B.6Find whole-number quotients and remainders with up to four-digit dividends (Dividing $2018$ by $6$ to get $336$ jumps with remainder $2$, locating $2018$ in the cycle.)4.OA.A.3Solve multi-step word problems using four operations with whole numbers (Combining the starting value $f(2)=1$ with the $336$ jumps of six to get $1 + 2016 = 2017$.)
⭐ 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.
⭐ 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.
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