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Difference between revisions of "Northeastern WOOTers Mock AIME I Problems/Problem 2"

(Created page with "== Problem 2 == It is given that <math>181^2</math> can be written as the difference of the cubes of two consecutive positive integers. Find the sum of these two integers....")
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Revision as of 21:56, 7 August 2021

Problem 2

It is given that $181^2$ can be written as the difference of the cubes of two consecutive positive integers. Find the sum of these two integers.



Solution

Let the smaller integer be $x$. Then (x+1)3−x3=18123x2+3x+1=18123x(x+1)=1812−13x(x+1)=(180)(182)x(x+1)=(60)(182) Since $x(x + 1) \approx x^2$ and $60 \cdot 182 \approx (60 \sqrt{3})^2 \approx 104^2$, we might guess $x = 104$. Through this method or others, we find that $x = 104$ and the sum of the two integers is $\boxed{209}$.

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