AMC 10 · 2009 · #24

Grade 11 algebra
Archetype Lattice / Grid Pattern from Small Cases └ Sub-archetype Figure-N Growth Sequence
logarithm-propertiesrecursive-sequenceexponential-function pattern-recognitionbound-inequality-then-enumerate ↑ Prerequisites: logarithm-properties
📏 Long solution 💡 4 insights
Problem
A tower of powers is built, and a huge number is formed from it. Count how many times a logarithm can be applied before it becomes impossible.

Pick an answer.

(A)
2009
(B)
2010
(C)
2011
(D)
2012
(E)
2013
How to solve
Strategy Change Focus / Count the Complement

Tool #16 (Change Focus): the size of B is hopeless to compute and completely beside the point. All that matters is which pair of consecutive towers B sits between, because log₂ carries the window (T(n), T(n+1)) exactly onto (T(n-1), T(n)). So the question "how many logs" becomes the question "which floor". Tool #4 (Introduce a Variable): naming t = T(2009) and u = T(2008), with the single relation t = 2^u, makes every later inequality one line long. Tool #7 (Identify Subproblems): the count splits into a general lemma about how deep a number's log-chain runs, and a separate placement question about where B lands. Tool #5 (Look for a Pattern): the one identity log₂ T(n) = T(n-1) is the entire engine — each log steps down exactly one floor. Tool #14 (Extreme Principle): the decisive case is the boundary. A number sitting exactly on a tower and a number sitting strictly above it have different depths, differing by exactly one, and that one unit separates two of the answer choices — so both inequalities have to be proved strict, not estimated. Tool #9 (Solve an Easier Related Problem): replacing 2009 by a small m gives numbers small enough to log by hand, which is how the final count gets checked independently — and it also exposes that the smallest case is degenerate.

1STEP 1

Say exactly what defined forbids

Each application is legal only while the input stays positive.

x₀ = B, x_j+1 = log₂ x_j; x_k is defined ⇔ x₀, x₁, …, x_k-1 ＞ 0
2STEP 2

One identity runs everything

One identity relates a tower to the one below it.

log₂ T(n) = T(n-1) (n ≥ 2); T(n) ＜ x ＜ T(n+1) → T(n-1) ＜ log₂ x ＜ T(n)
3STEP 3

Depth of an exact tower

That gives the depth of an exact tower.

T(n) → T(n-1) → … → T(1) = 2 → 1 → 0 (stop); D(T(n)) = (n-1) + 2 = n+1
4STEP 4

Strictly above a tower buys one more

Being strictly above a tower buys one more step.

T(n) ＜ x ＜ T(n+1) → x_n-1 ∈ (2,4) → x_n ∈ (1,2) → x_n+1 ∈ (0,1) → x_n+2 ＜ 0 (stop); D(x) = n+2
5STEP 5

Peel three logarithms off B

Peeling three logarithms off the number is straightforward.

x₁ = t^t u, x₂ = tu + T(2007), x₃ = log₂(tu + T(2007))
6STEP 6

Trap the result strictly between two towers

The result is trapped strictly between two towers.

T(2008) = log₂ t ＜ x₃ ＜ log₂(2t²) = 1 + 2u ＜ 2^u = T(2009)
7STEP 7

Add up the count

Adding the counts gives 2013, choice (E).

D(B) = 3 + D(x₃) = 3 + (2008 + 2) = 2013; equivalently T(2011) ＜ B ＜ T(2012) → D(B) = 2011 + 2 = 2013 (E)
Answer
2013
Run the same construction with 2009 replaced by m = 2, where every number is small enough to check by hand. Then T(2) = 4, A = 4⁴ = 256, and B = 4²⁵⁶ = 2⁵¹². Logging repeatedly gives 512 → 9 → log₂ 9 ≈ 3.170 → ≈ 1.664 → ≈ 0.735 → ≈ -0.444, and the next step is impossible. That is six logarithms, matching the general count m + 4. The placement argument agrees too: T(4) = 65536 ＜ 2⁵¹² ＜ T(5), so n = 4 and Step 4 gives D = 6. Both routes land on the same number in a case that can be verified digit by digit. Now the warning that makes this problem dangerous. The case m = 1 does not follow the pattern: there T(1) = 2, A = 4, and B = 2⁴ = 16, which is exactly T(3) — a tower, not a number strictly between two towers. Its depth is 16 → 4 → 2 → 1 → 0, that is D = 4, not 1 + 4 = 5. Anyone who reads the pattern off the very smallest case gets m + 3 and the wrong letter. The pattern m + 4 starts only at m = 2, and the reason is exactly the inequality used in Step 6: it needs T(m-1) ≥ 2, which has no meaning when m = 1. So the strictness in Step 4 is not a technicality; it is the only thing separating the right answer from 2012. That also identifies the intended trap: 2012 is precisely D(T(2011)), the answer obtained by rounding x₃ to T(2008) instead of proving it strictly larger, and it is also the answer obtained by wrongly demanding that the final output be positive. The smaller choices 2009, 2010, and 2011 come from stopping the descent one or more rungs above 0.
💡Key takeaway

Never try to compute a tower — just find which two towers your number sits between, because a number strictly above a tower survives exactly one more logarithm than the tower itself does.

  • Say exactly what defined forbids
  • One identity runs everything
  • Depth of an exact tower
  • Strictly above a tower buys one more
  • Peel three logarithms off B
  • Trap the result strictly between two towers
  • Add up the count

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