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Olympiad workbook
Exam
Olympiad workbook
Subject
IOQM
Chapter
Sequences and Series

Olympiad workbook · IOQM · Sequences and Series

Integers 1,2,3 . n where n>2 , are written on a board. Two numbers...

Integers 1,2,3…….n1,2,3 \ldots \ldots . \mathrm{n} where n>2\mathrm{n}>2, are written on a board. Two numbers m,k\mathrm{m}, \mathrm{k} such that 1<m<n.1<k<n1 < \mathrm{m} < \mathrm{n} .1 < \mathrm{k} < \mathrm{n} are removed and the average of the remaining numbers is found to be 17 . What is the maximum sum of the two removed numbers?

Correct answer

51

Step-by-step solution

n(n+1)2−(2n−1)n−2<17<n(n+1)2−3n−2n2+n−4n+22(n−2)<17<n2+n−62(n−2)n2+3n+22(n−2)<17<(n+3)(n−2)2(n−2)n−12<17<n+32n<35 and n>31n=32,33,34 case- 1,n=32\begin{aligned} & \frac{\frac{n(n+1)}{2}-(2 n-1)}{n-2} < 17 < \frac{\frac{n(n+1)}{2}-3}{n-2} \\ & \frac{n^2+n-4 n+2}{2(n-2)} < 17 < \frac{n^2+n-6}{2(n-2)} \\ & \frac{n^2+3 n+2}{2(n-2)} < 17 < \frac{(n+3)(n-2)}{2(n-2)} \\ & \frac{n-1}{2} < 17 < \frac{n+3}{2} \\ & n < 35 \text { and } n>31 \\ & n=32,33,34 \\ & \text { case- } 1, n=32\end{aligned} n(n+1)2−p(n−2)=17⇒n(n+1)2−17(n−2)=pp=18 case- 2,n=33⇒p=34 case- 3,n=34⇒p=51 Maximum sum =51\begin{aligned} & \frac{\frac{n(n+1)}{2}-p}{(n-2)}=17 \Rightarrow \frac{n(n+1)}{2}-17(n-2)=p \\ & p=18 \\ & \text { case- } 2, n=33 \Rightarrow p=34 \\ & \text { case- } 3, n=34 \Rightarrow p=51 \\ & \text { Maximum sum }=51\end{aligned}
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