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# Perfect square Problem | AIME I, 1999 | Question 3

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

Perfect Square

Integers

Inequalities

## Check the Answer

Answer: is 38.

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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