Northeastern WOOTers Mock AIME I Problems/Problem 2
Problem 2
It is given that 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 . 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
and
, we might guess
. Through this method or others, we find that
and the sum of the two integers is
.