Problem:
Consider
A=log(2013+log(2012+log(2011+log(⋯+log(3+log2)⋯))))
Which of the following intervals contains A ?
Answer Choices:
A. (log2016,log2017)
B. (log2017,log2018)
C. (log2018,log2019)
D. (log2019,log2020)
E. (log2020,log2021)
Solution:
Let An=log(n+log((n−1)+log(⋯+log(3+log2)⋯))). Note that 0< log2=A2<1. If 0<Ak−1<1, then k<k+Ak−1<k+1. Hence 0<logk<log(k+Ak−1)=Ak<log(k+1)≤1, as long as logk>0 and log(k+1)≤1, which occurs when 2≤k≤9. Thus 0<An<1 for 2≤n≤9.
Because 0<A9<1, it follows that 10<10+A9<11, and so 1=log(10)< log(10+A9)=A10<log(11)<2. If 1<Ak−1<2, then k+1<k+Ak−1< k+2. Hence 1<log(k+1)<log(k+Ak−1)=Ak<log(k+2)≤2, as long as log(k+1)>1 and log(k+2)≤2, which occurs when 10≤k≤98. Thus 1<An<2 for 10≤n≤98.
In a similar way, it can be proved that 2<An<3 for 99≤n≤997, and 3<An<4 for 998≤n≤9996.
For n=2012, it follows that 3<A2012<4, so 2016<2013+A2012<2017 and log2016<A2013<log2017.
The problems on this page are the property of the MAA's American Mathematics Competitions