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

\begin{align*} (x + 1)^3 - x^3 &= 181^2 \\ 3x^2 + 3x + 1 &= 181^2 \\ 3x(x + 1) &= 181^2 - 1 \\ 3x(x + 1) &= (180)(182) \\ x(x + 1) &= (60)(182) \end{align*}

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 $\boxed{209}$ .

\begin{align*} ((2x+3)^3)' &= 3(2x+3)^2 \cdot (2x+3)' \\ &= 3(2x+3)^2 \cdot 2 \\ &= 6(2x+3)^2. \end{align*}

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Northeastern WOOTers Mock AIME I Problems/Problem 2

Problem 2

Solution