1***f(13)=4
A)
n^2-n-2008=0
44*2=88
28 terms left
\text{Let f(n) be the integer closest to} \ {\sqrt{n}}\\ \text{find the value of }\\ \sum_{i=1}^{2008}{\frac{1}{f(i)}}\\ \text{NOTE: f(6=2),f(7)=3,f(12)=3,f(13)=2 etc.}\\ (A)88\frac{28}{45} \ \ (B)90\frac{3}{5} \ \ (C)87\frac{5}{6} \ \ (D)89\frac{1}{5} %
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3 Answers
1got the answer as A
@che ramanujan contest is conducted by the association mathematics teacher of india this was just a question from screening test at INTER level
it is conducted in tamil nadu
http://en.wikipedia.org/wiki/National_Mathematics_Talent_Contest