AMC 10 · 2019 · #22

Grade 8 algebra
recursive-sequencesequences-geometricexponentspolynomial-factoringestimation pattern-recognitionidentify-subproblems ↑ Prerequisites: recursive-sequencesequences-geometricexponents
📏 Long solution 💡 4 insights
Problem
A sequence starts at 5, and each next term is the previous term squared plus five times it plus four, all divided by the previous term plus six. Let m be the smallest index at which the term drops to four plus two to the power minus twenty or below. Find which interval m lies in.

Pick an answer.

(A)
[9,26]
(B)
[27,80]
(C)
[81,242]
(D)
[243,728]
(E)
$[729,\infty)$
How to solve
Strategy Solve an Easier Related Problem

Tool #13 (Algebra): factor the numerator to expose the fixed point x = 4, then substitute y_n = x_n - 4 to move the equilibrium to 0. Tool #9 (Easier Problem): near y = 0 the recursion becomes nearly linear (y_n+1 ≈ 9/10 y_n) — a geometric sequence we can solve in closed form. Tool #5 (Pattern): the ratio 9/10 together with the wide answer-choice intervals lets us estimate m by m ≈ 20 log 2 / log(10/9). Tool #6 (Guess and Check): plug m ≈ 130 into the answer-choice intervals. Tool #3 (Eliminate): the wide intervals are robust to crude estimates.

1STEP 1

Factor the numerator

Factoring shows four is a fixed point.

x_n+1 = (x_n+1)(x_n+4)/(x_n+6); x = 4 is fixed
2STEP 2

Rewrite in terms of the gap

Rewrite using the gap from four.

y_n+1 = (y_n (y_n + 9))/(y_n + 10)
3STEP 3

How fast the gap shrinks

Each step multiplies the gap by about nine tenths.

y_n+1 ≈ 9/10 y_n → y_n ≈ (9/10)ⁿ
4STEP 4

Estimate the number of steps

Logarithms estimate the step count.

m ≈ (20 log₁₀ 2)/log₁₀(10/9)
5STEP 5

Evaluate numerically

That comes out around one hundred thirty.

m ≈ 6.02/0.046 ≈ 131
6STEP 6

Pick the interval

Choose the interval containing it.

131 ∈ [81, 242]
7STEP 7

Read the answer

The answer is eighty-one through two hundred forty-two.

[81, 242]
Answer
[81,242]
Two sanity checks. (1) The actual ratio (y_n + 9)/(y_n + 10) starts at 10/11 ≈ 0.909 (when y₀ = 1) and tends to 9/10 = 0.9. So in the early phase the decay is slightly slower than 9/10 per step, meaning the true m is a touch larger than the (9/10)^m = 2⁻²⁰ estimate — still in [81, 242] with room. (2) The bounds: if we replaced the ratio with 10/11 throughout, we would need m log(11/10) ≥ 20 log 2, i.e. m ≥ 20 · 0.301 / 0.0414 ≈ 145 — still in [81, 242]. The lower bound on m would be at least 20 · 0.301 / 0.046 ≈ 131 once we are deep in the small-y regime. The intervals are wide enough that any reasonable estimate lands in (C).
💡Key takeaway

This AMC 12 problem only needs Grade 8 logs and geometric sequences — shift coordinates with y_n = x_n - 4 to expose the ratio 9/10, then (9/10)^m ≈ 1/2²⁰ gives m ≈ 131, which lands in [81, 242].