AMC 10 · 2014 · #24

Grade 9 algebracounting
absolute-valuerecursive-sequencefunction-compositionparity work-backwardscaseworksystematic-enumeration ↑ Prerequisites: absolute-valuerecursive-sequence
📏 Medium solution 💡 4 insights
Problem
One rule is applied a hundred times and the result must land exactly on zero. Count the starting values.

Pick an answer.

(A)
299
(B)
300
(C)
301
(D)
302
(E)
303
How to solve
Strategy Work Backwards

Nobody can write out f₁₀₀ as a formula, so the plan splits the problem in two along its natural seam. Tool #9 (Solve an Easier Related Problem) replaces 'iterate a whole function 100 times' with 'iterate a single number 100 times under t↦ |t|-1', because after the first step the outcome depends only on the one number f₀(x). Tool #11 (Work Backwards) then runs that easier problem in reverse: instead of guessing what f₁₀₀ looks like from a picture of f₁,f₂,f₃, it asks which starting numbers are still sitting on 0 after exactly 100 steps, and answers that question with proof — the starting number must be an even whole number of size at most 100. Tool #7 (Identify Subproblems) then unfolds f₀ into three straight-line pieces by splitting the number line at x=-100 and x=100, so each allowed target value can be hit at most once per piece. Tool #2 (Make a Systematic List) finishes: list the allowed target values, count how many x each one has, and add. Counting the exact preimages beats extrapolating a pattern from a graph, because the whole difficulty of this problem lives in the two endpoints where two of the three pieces collide.

1STEP 1

Unfold f₀ into three lines

The first function is three straight pieces.

f₀(x)= x+200, & x ≤ -100 ; -x, & -100 ≤ x ≤ 100 ; x-200, & x ≥ 100
2STEP 2

Follow one number through the rule

Following one number shows what survives.

b=m+r, 0 ≤ r < 1: after m steps the value is r; r > 0→ tail cycles r-1,-r (never 0); r=0→ tail cycles 0,-1
3STEP 3

Turn f₁₀₀(x)=0 into a rule for f₀

The target becomes a short list of first values.

f₁₀₀(x)=0⇔ |f₀(x)|∈{0,2,4,…,100}⇔ f₀(x)∈{0,± 2,± 4,…,± 100}
4STEP 4

Solve f₀(x)=v on each piece

Each value is solved on each piece.

x=v-200, x=-v, x=v+200; |v| < 100→ 3 distinct roots, v=± 100→ 2 distinct roots
5STEP 5

Count the targets and add

Adding the counts gives 301, choice (C).

99 · 3+2 · 2=297+4=301 → (C)
Answer
301
Spot checks confirm the rule found in step 3. At x=0, f₀(0)=0+100-100=0, an allowed even target, and the values run 0,-1,0,-1,…, landing on 0 at every even step, so f₁₀₀(0)=0. At x=1, f₀(1)=-1, which is odd, and the values run -1,0,-1,0,…, so f₁₀₀(1)=-1 ≠ 0. At x=302, f₀(302)=102 > 100, and 100 steps only bring it down to 2, so it fails. At the extremes x=-300 and x=300 we get f₀=-100 and f₀=100, both allowed, so the solution set really does stretch from -300 to 300. The count itself is close to but not equal to 300: the tempting wrong move is (300-(-300))/2=300, which forgets that a list with both endpoints included has one more member than the number of gaps. The answer choices 299 through 303 are exactly this family of off-by-one traps, and 301 is the one that survives.
💡Key takeaway

The rule 'take the size, then subtract 1' walks a number down to 0 one step at a time, so after 100 steps you are on 0 only if you started at an even whole number no bigger than 100 — then count the x that produce those starts: three each, except at the peak and the valley.

  • Unfold f₀ into three lines
  • Follow one number through the rule
  • Turn f₁₀₀(x)=0 into a rule for f₀
  • Solve f₀(x)=v on each piece
  • Count the targets and add