Northeastern WOOTers Mock AIME I Problems/Problem 2

Revision as of 20:57, 7 August 2021 by Skyguy88 (talk | contribs) (Solution)

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)3x3=18123x2+3x+1=18123x(x+1)=181213x(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 $\boxed{209}$.

((2x+3)3)=3(2x+3)2(2x+3)=3(2x+3)22=6(2x+3)2.