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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....")
 
m (→‎Solution)
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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 \\
Line 16: Line 17:
 
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 $181^2$ 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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