AMC 10 · 2019 · #23

Grade 11 algebra
Archetype Telescoping Collapse └ Sub-archetype Product Ratio Cancellation
logarithm-propertiesrecursive-sequencetelescoping-sumestimation convert-to-algebrapattern-recognitionestimation ↑ Prerequisites: logarithm-propertiesrecursive-sequence
📏 Medium solution 💡 3 insights
Problem
Two invented operations are built from base-seven logarithms: one raises a number to the logarithm of another, and the other raises it to the reciprocal of that logarithm. A sequence starts at its third term and every later term is made from the one before it by applying both operations. Find the base-seven logarithm of the 2019th term, rounded to the nearest integer.

Pick an answer.

(A)
8
(B)
9
(C)
10
(D)
11
(E)
12
How to solve
Strategy Organize Information in More Ways

Computing a₂₀₁₉ directly is hopeless: each step stacks another exponent on top of an already enormous tower. But the question does not ask for a₂₀₁₉, it asks for log₇(a₂₀₁₉), which is a strong hint to stop tracking the terms and start tracking their logarithms. Rewriting both invented symbols in log form turns them into ordinary multiplication and division, so I introduce b_n = log₇(a_n) and rebuild the recursion in that variable. Two small cases show what the new recursion does, the pattern they suggest is confirmed by a telescoping product, and the final estimate only needs 2019 pinned between two powers of 2, which also rules out four of the five choices.

1STEP 1

Read both symbols in log-land

Read both symbols in log-land.

log₇(a ◆ b) = (log₇ a)(log₇ b), log₇(a ♥ b) = (log₇ a)/(log₇ b)
2STEP 2

Name the log of each term

Name each term's logarithm.

b_n = (log₇ n)/(log₇ (n-1)) b_n-1 = b_n-1 log_n-1 n
3STEP 3

Start the chain, then test small n

A few small terms reveal the pattern.

b₃ = log₂ 3, b₄ = log₂ 3 · log₃ 4 = log₂ 4 = 2, b₅ = log₂ 4 · log₄ 5 = log₂ 5
4STEP 4

Cancel the whole chain

The whole chain cancels through.

b_n = (log 3)/(log 2) · (log 4)/(log 3) · (log 5)/(log 4) … (log n)/(log (n-1)) = (log n)/(log 2) = log₂ n
5STEP 5

Trap the value between powers of 2

Trapping it between powers of two gives 11.

2¹⁰ = 1024 ＜ 2019 ＜ 2048 = 2¹¹ → 10 ＜ log₂ 2019 ＜ 11, log₂ 2019 ≈ 10.98
Answer
11
The closed form b_n = log₂ n is checked at both ends of the argument. At the start it returns b₃ = log₂ 3, which is exactly what a₃ = 3¹/(log₇ 2) gives directly, and it reproduces the hand-computed b₄ = 2 and b₅ = log₂ 5. The size of the answer is also sensible: base-2 logarithms of four-digit numbers sit near 10 or 11, and all five choices lie in the band 8 to 12, so the contest is testing which base survives the cancellation. Base 7 would have given log₇ 2019 ≈ 3.9 and base 10 would have given about 3.3, neither of which is offered, while base 2 lands squarely inside the choice list. A direct evaluation confirms log₂ 2019 ≈ 10.979, whose nearest integer is 11, choice (D).
💡Key takeaway

When a problem hides its numbers up in exponents, take the logarithm of everything: towers turn into products, the products cancel down a chain, and reaching index 2019 costs no more work than reaching index 4.

  • Read both symbols in log-land
  • Name the log of each term
  • Start the chain, then test small n
  • Cancel the whole chain
  • Trap the value between powers of 2

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