Difference between revisions of "2003 AIME II Problems/Problem 10"
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== Problem == | == Problem == | ||
| + | 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 == | ||
| + | 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}} | ||
| + | |||
| + | [[Category: Intermediate Algebra Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 13:14, 25 June 2021
Problem
Two positive integers differ by 
. 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 
and 
, so we have 
. Square both sides to get 
. Thus, 
must be a square, so we have 
, and 
. The sum of these two factors is 
, so they must both be even. To maximize , we want to maximixe 
, so we let it equal 
and the other factor 
, but solving gives 
, 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 
and the other 
, which gives 
. This checks, so the solution is 
.
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 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
