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Difference between revisions of "2003 AIME II Problems/Problem 10"

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== Problem ==
 
== Problem ==
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Two positive integers differ by <math>60.</math> The sum of their square roots is the square root of an integer that is not a perfect square. What is the maximum possible sum of the two integers?
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Two positive integers differ by <math>60</math>. The sum of their square roots is the square root of an integer that is not a perfect square. What is the maximum possible sum of the two integers?
  
 
== Solution ==
 
== Solution ==
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{{solution}}
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Call the two integers <math>b</math> and <math>b+60</math>, so we have <math>\sqrt{b}+\sqrt{b+60}=\sqrt{c}</math>. Square both sides to get <math>2b+60+2\sqrt{b^2+60b}=c</math>. Thus, <math>b^2+60b</math> must be a square, so we have <math>b^2+60b=n^2</math>, and <math>(b+n+30)(b-n+30)=900</math>. The sum of these two factors is <math>2b+60</math>, so they must both be even. To maximize <math>b</math>, we want to maximixe <math>b+n+30</math>, so we let it equal <math>450</math> and the other factor <math>2</math>, but solving gives <math>b=196</math>, which is already a perfect square, so we have to keep going. In order to keep both factors even, we let the larger one equal <math>150</math> and the other <math>6</math>, which gives <math>b=48</math>. This checks, so the solution is <math>48+108=\boxed{156}</math>.
  
 
== See also ==
 
== See also ==
 +
Video Solution from Khan Academy:
 +
https://www.youtube.com/watch?v=Hh3iY4tdkGI
 
{{AIME box|year=2003|n=II|num-b=9|num-a=11}}
 
{{AIME box|year=2003|n=II|num-b=9|num-a=11}}
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[[Category: Intermediate Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 13:14, 25 June 2021

Problem

Two positive integers differ by $60$. The sum of their square roots is the square root of an integer that is not a perfect square. What is the maximum possible sum of the two integers?

Solution

Call the two integers $b$ and $b+60$, so we have $\sqrt{b}+\sqrt{b+60}=\sqrt{c}$. Square both sides to get $2b+60+2\sqrt{b^2+60b}=c$. Thus, $b^2+60b$ must be a square, so we have $b^2+60b=n^2$, and $(b+n+30)(b-n+30)=900$. The sum of these two factors is $2b+60$, so they must both be even. To maximize $b$, we want to maximixe $b+n+30$, so we let it equal $450$ and the other factor $2$, but solving gives $b=196$, which is already a perfect square, so we have to keep going. In order to keep both factors even, we let the larger one equal $150$ and the other $6$, which gives $b=48$. This checks, so the solution is $48+108=\boxed{156}$.

See also

Video Solution from Khan Academy: https://www.youtube.com/watch?v=Hh3iY4tdkGI

2003 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 9 		Followed by
Problem 11
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All AIME Problems and Solutions

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