Try this beautiful problem from the Pre-RMO, 2017 based on Average and Integers.

## Average and Integer – PRMO 2017

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.

- is 107
- is 51
- is 840
- cannot be determined from the given information

**Key Concepts**

Average

Integers

Maximum

## Check the Answer

But try the problem first…

Answer: is 51.

PRMO, 2017, Question 15

Elementary Algebra by Hall and Knight

## Try with Hints

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.

## Other useful links

- https://www.cheenta.com/rational-number-and-integer-prmo-2019-question-9/
- https://www.youtube.com/watch?v=lBPFR9xequA

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