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2024 AMC 10B Problem 13

Problem 13 of 25IntermediateAlgebraNumber Theory

Positive integers xx and yy satisfy the equation x+y=1183.\sqrt{x} + \sqrt{y} = \sqrt{1183}. What is the minimum possible value of x+y?x + y?

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Solution

Since 1183=7132,1183 = 7 \cdot 13^2, we have 1183=137.\sqrt{1183} = 13\sqrt7. Squaring the given equation shows that xy\sqrt{xy} is rational. Thus xx and yy have the same squarefree part: write x=da2x=da^2 and y=db2,y=db^2, where dd is squarefree and a,ba,b are positive integers. Then (a+b)d=137,(a+b)\sqrt d=13\sqrt7, so d=7d=7 and a+b=13.a+b=13. Therefore x+y=7(a2+b2),x+y=7(a^2+b^2), which is smallest when aa and bb are as close as possible. Taking a=6a=6 and b=7b=7 gives x+y=7(36+49)=595.x+y=7(36+49)=595. Thus, B is the correct answer.

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Concepts: radical · prime factorization · optimization

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.