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# Average and Integers | PRMO 2017 | Question 15

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

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.

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