AMC 10 · 2017 · #7

Grade 4 algebra
Archetype Lattice / Grid Pattern from Small Cases └ Sub-archetype Figure-N Growth Sequence
recursive-sequenceparitysequences-arithmetic work-backwardspattern-recognition ↑ Prerequisites: recursive-sequenceparity
📏 Medium solution 💡 2 insights
Problem
Two rules define a function, and one of them applies to the wanted input. Find that value.

Pick an answer.

(A)
2017
(B)
2018
(C)
4034
(D)
4035
(E)
4036
How to solve
Strategy Work Backwards

A recursive definition only tells you how one value depends on earlier ones, so the value at 2017 is unreachable going forward — you would have to compute two thousand values first. Tool #11 (Work Backwards) is the natural fit: start at f(2017) and keep replacing it by the earlier value the rule points to, until the chain hits the one value we are handed outright, f(1) = 2. Tool #5 (Look for a Pattern) then does the real work, because the chain turns out to take the same size step every time. Tool #9 (Solve an Easier Related Problem) lets us rehearse the counting on a chain short enough to check by hand, and Tool #2 (Make a Systematic List) turns the full chain of indices into something countable instead of something we have to walk through.

1STEP 1

See which rule 2017 triggers

The wanted input triggers the second rule.

2017 odd, 2017 ＞ 1 → f(2017) = f(2015) + 2
2STEP 2

The chain never leaves the odd numbers

The chain stays on the odd numbers.

2017 → 2015 → 2013 → … → 3 → 1, f(1) = 2
3STEP 3

Rehearse the count on a short chain

A short chain shows how to count the steps.

5 → 3 → 1: (5-1)/2 = 2 arrows, f(5) = f(1) + 2 · 2 = 6
4STEP 4

Count the arrows down from 2017

There are 1008 steps down to the start.

(2017 - 1)/2 = 2016/2 = 1008 arrows
5STEP 5

Add up every +2

Adding them gives 2018, choice (B).

f(2017) = f(1) + 2 · 1008 = 2 + 2016 = 2018 → (B)
Answer
2018
Check the size first. Along the odd numbers the index climbs by 2 while the value climbs by 2, so the value grows at about the same rate as n — it should land near 2017, not near double it. That immediately makes (C) 4034 and (E) 4036 look wrong: those are roughly 2 · 2017, which is what you get if you wrongly add 2 for each single increase in n instead of for each increase of 2. Next check the landing: 1008 arrows starting at 2017 end at 2017 - 2 · 1008 = 1, exactly the base case, so nothing is left over and nothing is double-counted. Finally the small case computed directly agrees: f(3) = f(1) + 2 = 4 and f(5) = f(3) + 2 = 6, matching the recipe. The value 2018 is choice (B).
💡Key takeaway

When a rule sends 2017 back to 2015, the chain can only ever land on odd numbers — so count how many steps it takes to reach the base case, then multiply by what each step adds.

  • See which rule 2017 triggers
  • The chain never leaves the odd numbers
  • Rehearse the count on a short chain
  • Count the arrows down from 2017
  • Add up every +2

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