AMC12 2013 A

AMC12 2013 A · Q21

AMC12 2013 A · Q21. It mainly tests Logarithms (rare), Algebra misc.

Consider $A = \log \left(2013 + \log \left(2012 + \log \left(2011 + \log(\cdots + \log (3 + \log 2) \cdots)\right)\right)\right)$. Which of the following intervals contains $A$?

考虑 $A = \log \left(2013 + \log \left(2012 + \log \left(2011 + \log(\cdots + \log (3 + \log 2) \cdots)\right)\right)\right)$。以下哪个区间包含 $A$？

(A) $(\log 2016, \log 2017)$ $(\log 2016, \log 2017)$

(B) $(\log 2017, \log 2018)$ $(\log 2017, \log 2018)$

(D) $(\log 2019, \log 2020)$ $(\log 2019, \log 2020)$

(E) $(\log 2020, \log 2021)$ $(\log 2020, \log 2021)$

Answer

Correct choice: (A)

正确答案：（A）

Solution

Answer (A): Let $A_n=\log\!\bigl(n+\log((n-1)+\log(\cdots+\log(3+\log 2)\cdots))\bigr)$. Note that $0<\log 2=A_2<1$. If $0<A_{k-1}<1$, then $k<k+A_{k-1}<k+1$. Hence $0<\log k<\log(k+A_{k-1})=A_k<\log(k+1)\le1$, as long as $\log k>0$ and $\log(k+1)\le1$, which occurs when $2\le k\le9$. Thus $0<A_n<1$ for $2\le n\le9$. Because $0<A_9<1$, it follows that $10<10+A_9<11$, and so $1=\log(10)<\log(10+A_9)=A_{10}<\log(11)<2$. If $1<A_{k-1}<2$, then $k+1<k+A_{k-1}<k+2$. Hence $1<\log(k+1)<\log(k+A_{k-1})=A_k<\log(k+2)\le2$, as long as $\log(k+1)>1$ and $\log(k+2)\le2$, which occurs when $10\le k\le98$. Thus $1<A_n<2$ for $10\le n\le98$. In a similar way, it can be proved that $2<A_n<3$ for $99\le n\le997$, and $3<A_n<4$ for $998\le n\le9996$. For $n=2012$, it follows that $3<A_{2012}<4$, so $2016<2013+A_{2012}<2017$ and $\log 2016<A_{2013}<\log 2017$.

答案（A）：令 $A_n=\log\!\bigl(n+\log((n-1)+\log(\cdots+\log(3+\log 2)\cdots))\bigr)$。注意到 $0<\log 2=A_2<1$。若 $0<A_{k-1}<1$，则 $k<k+A_{k-1}<k+1$。因此在 $\log k>0$ 且 $\log(k+1)\le1$ 的条件下，有 $0<\log k<\log(k+A_{k-1})=A_k<\log(k+1)\le1$。而当 $2\le k\le9$ 时上述条件成立。于是当 $2\le n\le9$ 时，$0<A_n<1$。因为 $0<A_9<1$，可得 $10<10+A_9<11$，从而 $1=\log(10)<\log(10+A_9)=A_{10}<\log(11)<2$。若 $1<A_{k-1}<2$，则 $k+1<k+A_{k-1}<k+2$。因此在 $\log(k+1)>1$ 且 $\log(k+2)\le2$ 的条件下，有 $1<\log(k+1)<\log(k+A_{k-1})=A_k<\log(k+2)\le2$。而当 $10\le k\le98$ 时上述条件成立。于是当 $10\le n\le98$ 时，$1<A_n<2$。同理可证：当 $99\le n\le997$ 时 $2<A_n<3$；当 $998\le n\le9996$ 时 $3<A_n<4$。当 $n=2012$ 时，可得 $3<A_{2012}<4$，所以 $2016<2013+A_{2012}<2017$，并且 $\log 2016<A_{2013}<\log 2017$。

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