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Algebra Arithmetic Math Olympiad PRMO

Average and Integers | PRMO 2017 | Question 15

Try this beautiful problem from the Pre-RMO, 2017 based on Average and Integers. You may use sequential hints to solve the problem.

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.

Source
Suggested Reading

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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