Try this beautiful problem from the Pre-RMO, 2017 based on Average and Integers.
Integers 1,2,3,....,n where n>2, are written on a board. Two numbers m,k such that 1<m<n,1<k<n are removed and the average of the remaining numbers is found to be 17, find the maximum sum of the two removed numbers.
Average
Integers
Maximum
But try the problem first...
Answer: is 51.
PRMO, 2017, Question 15
Elementary Algebra by Hall and Knight
First hint
\(\frac{{n(n+1)}{2}-(2n-1)}{n-2}\)<17<\(\frac{{n(n+1)}{2}-3}{n-2}\)
\(\Rightarrow \frac{n^{2}+n-4n+2}{2(n-2)}<17<\frac{n^{2}+n+6}{2(n-2)}\)
\(\Rightarrow \frac{n-1}{2}<17<\frac{n+3}{2}\)
Second Hint
\(\Rightarrow n<35 and n>31\)
\(\Rightarrow\) n=32,33,34
Final Step
First case n=32 \(\frac{\frac{n(n+1)}{2}-p}{(n+2)}\)=17
\(\Rightarrow\) p=18
Second case, n=33 \(\Rightarrow\) p=34
Third case, n=34 \(\Rightarrow\) p=51.
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