2013 AMC 12A Problem 21

Problem 21 of 25HarderAlgebra

Consider

\begin{gathered} A = \\ \tiny \log(2013 + \log(2012 + \log(2011 + \log(\cdots + \log(3 + \log 2)\cdots)))). \end{gathered}

Which of the following intervals contains

A?

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Solution

Let

\tiny A_n = \log(n + \log((n-1) + \cdots + \log(3 + \log 2)\cdots)).

One checks

0 \lt A_n \lt 1

for

2 \le n \le 9,

then

1 \lt A_n \lt 2

for

10 \le n \le 98,

then

2 \lt A_n \lt 3

for

99 \le n \le 997,

and

3 \lt A_n \lt 4

for

998 \le n \le 9996.

Hence

3 \lt A_{2012} \lt 4,

so

2016 \lt 2013 + A_{2012} \lt 2017

and therefore

\log 2016 \lt A \lt \log 2017.

Thus, the correct answer is A.

More practice

Concepts: logarithm · bounding to limit cases · induction

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.