AMC 10 · 2013 · #21

Grade 11 algebra
Archetype Small-Domain Constrained Search └ Sub-archetype Logic-Deduction / Elimination
logarithm-propertiesrecursive-sequenceestimation convert-to-algebraextreme-principle ↑ Prerequisites: logarithm-properties
📏 Medium solution 💡 3 insights
Problem
Logarithms are nested thousands of layers deep. Decide which interval the value lands in.

Pick an answer.

(A)
$(\log 2016, \log 2017)$
(B)
$(\log 2017, \log 2018)$
(C)
$(\log 2018, \log 2019)$
(D)
$(\log 2019, \log 2020)$
(E)
$(\log 2020, \log 2021)$
How to solve
Strategy Introduce a Variable

The nest is over two thousand layers deep, so unfolding it is not an option. But the outermost layer is already in plain sight: A = log(2013 + T), where T is the entire rest of the nest. So I name T with a letter. Because log preserves order, the five intervals turn into five unit-wide ranges for T, and the whole problem becomes: which two consecutive whole numbers trap T? Then I bound T from below and from above, keeping both bounds strict, since back-to-back choices leave no room for slack.

1STEP 1

Write the nest as a recursion

The nest is one plain recursion.

f(2)=log 2, f(n)=log(n+f(n-1)) (n ≥ 3), A=f(2013)=log(2013+f(2012))
2STEP 2

Turn the choices into a range for the tail

The choices become a range for the inner value.

T=f(2012), A=log(2013+T), log(2013+k) ＜ A ＜ log(2014+k)⇔ k ＜ T ＜ k+1
3STEP 3

Trap the tail from below

Positivity traps it from below.

f(n) ＞ 0 for all n ≥ 2 ⟹ T=log(2012+f(2011)) ＞ log 2012 ＞ log 1000=3
4STEP 4

Prove the tail never reaches 4

Induction traps it from above.

3 ≤ n ≤ 9996 and f(n-1) ＜ 4 ⟹ n+f(n-1) ＜ 9996+4=10⁴ ⟹ f(n)=log(n+f(n-1)) ＜ 4; 2012 ≤ 9996 ⟹ T ＜ 4
5STEP 5

Close the interval

That names the first interval, choice (A).

3 ＜ T ＜ 4 ⟹ 2016 ＜ 2013+T ＜ 2017 ⟹ log 2016 ＜ A ＜ log 2017
Answer
(log 2016, log 2017)
Two checks. Self-consistency: T should satisfy T = log(2012 + T). Trying T = 3.30 gives log(2015.30) = 3.3045, which reproduces itself, so T sits near 3.3045 — well inside (3, 4). That places A near log(2016.30), which is inside (log 2016, log 2017) as claimed. Stability: the argument never used the value of f(2011) or anything deeper, only that it is positive and below 4. Swapping the whole two-thousand-layer tail for any number between 0 and 4 moves 2013 + T by less than one unit, so no further unfolding can push A out of (A). It also matters that all the work was done inside the logs: log 2016 and log 2017 differ by only about 0.0002, so a decimal estimate of A itself would have been far too crude, while the gap between 2016 and 2017 is a full unit.
💡Key takeaway

Logs squash numbers so hard that a two-thousand-layer tower is still worth less than 4, so trap the inside between two whole numbers and the outside lands in exactly one interval.

  • Write the nest as a recursion
  • Turn the choices into a range for the tail
  • Trap the tail from below
  • Prove the tail never reaches 4
  • Close the interval

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