Northeastern WOOTers Mock AIME I Problems/Problem 2

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 \[(x + 1)^3 - x^3 = 181^2 \Rightarrow 3x(x + 1) = 181^2 - 1 \Rightarrow x(x + 1) = (60)(182).\] 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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