Consider A = log(2013 + log(2012 + log(2011 + log(… + log(3 + log 2)…)))). Which of the following intervals contains A?

2013 AMC 12A Problem 21

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21.

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?

(\log 2016, \log 2017)

(\log 2017, \log 2018)

(\log 2018, \log 2019)

(\log 2019, \log 2020)

(\log 2020, \log 2021)

Track where the nested value sits:

0 \lt A_n \lt 1

for small

n,

then

1 \lt A_n \lt 2,

etc.

For

n = 2012

the value satisfies

3 \lt A_{2012} \lt 4,

so estimate

2013 + A_{2012}

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,

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

and therefore

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

Thus, the correct answer is A.