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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Let the smaller integer be <math>x</math>. Then | Let the smaller integer be <math>x</math>. Then | ||
+ | |||
\begin{align*} | \begin{align*} | ||
(x + 1)^3 - x^3 &= 181^2 \ | (x + 1)^3 - x^3 &= 181^2 \ | ||
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x(x + 1) &= (60)(182) | x(x + 1) &= (60)(182) | ||
\end{align*} | \end{align*} | ||
+ | |||
Since <math>x(x + 1) \approx x^2</math> and <math>60 \cdot 182 \approx (60 \sqrt{3})^2 \approx 104^2</math>, we might guess <math>x = 104</math>. Through this method or others, we find that <math>x = 104</math> and the sum of the two integers is <math>\boxed{209}</math>. | Since <math>x(x + 1) \approx x^2</math> and <math>60 \cdot 182 \approx (60 \sqrt{3})^2 \approx 104^2</math>, we might guess <math>x = 104</math>. Through this method or others, we find that <math>x = 104</math> and the sum of the two integers is <math>\boxed{209}</math>. | ||
+ | |||
+ | \begin{align*} | ||
+ | ((2x+3)^3)' &= 3(2x+3)^2 \cdot (2x+3)' \ | ||
+ | &= 3(2x+3)^2 \cdot 2 \ | ||
+ | &= 6(2x+3)^2. | ||
+ | \end{align*} |
Revision as of 20:57, 7 August 2021
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
Since and , we might guess . Through this method or others, we find that and the sum of the two integers is .