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AIME I Algebra Arithmetic Math Olympiad USA Math Olympiad

Perfect square Problem | AIME I, 1999 | Question 3

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 2011 based on Rectangles and sides.

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1999 based on Perfect square and Integers.

Perfect square Problem – AIME I, 1999


Find the sum of all positive integers n for which \(n^{2}-19n+99\) is a perfect square.

  • is 107
  • is 38
  • is 840
  • cannot be determined from the given information

Key Concepts


Perfect Square

Integers

Inequalities

Check the Answer


But try the problem first…

Answer: is 38.

Source
Suggested Reading

AIME I, 1999, Question 3

Elementary Number Theory by David Burton

Try with Hints


First hint

\((n-10)^{2}\) \(\lt\) \(n^{2}-19n+99\) \(\lt\) \((n-8)^{2}\) and \(n^{2}-19n+99\) is perfect square then \(n^{2}-19n+99\)=\((n-9)^{2}\) that is n=18

Second Hint

and \(n^{2}-19n+99\) also perfect square for n=1,9,10

Final Step

then adding 1+9+10+18=38.

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