Let $A = \cfrac{1}{\cfrac{1}{2011}+\cfrac{1}{2012}+\cfrac{1}{2013}+\cfrac{1}{2014}+\cfrac{1}{2015}+\cfrac{1}{2016}}$. $B$ is the largest integer so that $B \le A$. Find $B$
I tried to use AM-GM inequality in the denominator but it is impossible
Asked
Active
Viewed 92 times
-1
-
4Here's a calculator – amrsa Jan 07 '20 at 12:02
-
5Big Hint: $2011 < 6A < 2016$ (why?) so $335 < A < 336,$ making $B=335$ – Brian Moehring Jan 07 '20 at 12:02
-
@Shaun I've edit that – liszt16 Jan 07 '20 at 12:04
-
@amrsa I don't need the answer, just how to solve it – liszt16 Jan 07 '20 at 12:05
-
1@TranTu Follow BrianMoehring's lead... – amrsa Jan 07 '20 at 12:06
-
@MartinR this is a duplicate, what do I do? Sorry for not searching question – liszt16 Jan 07 '20 at 13:13
1 Answers
6
$$ \frac{1}{\frac{1}{2011} + \frac{1}{2011} + \frac 1{2011} + \frac{1}{2011}+\frac{1}{2011}+\frac{1}{2011}} < \frac 1{\frac{1}{2011} +\frac{1}{2012} +\frac{1}{2013} +\frac{1}{2014} +\frac{1}{2015} +\frac{1}{2016}}\\ < \frac{1}{\frac 1{2016}+\frac{1}{2016}+\frac{1}{2016}+\frac{1}{2016}+\frac{1}{2016}+\frac{1}{2016}} $$
LHS $= \frac {2011}{6} > 335$ , RHS $= \frac {2016} {6} = 336$
liszt16
- 423
Sarvesh Ravichandran Iyer
- 76,105