Difference between revisions of "Northeastern WOOTers Mock AIME I Problems/Problem 2"
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Let the smaller integer be <math>x</math>. Then | Let the smaller integer be <math>x</math>. Then | ||
| − | + | <cmath> (x + 1)^3 - x^3 = 181^2 \Rightarrow 3x(x + 1) = 181^2 - 1 \Rightarrow x(x + 1) = (60)(182). </cmath> | |
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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>. | ||
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Latest revision as of 21:02, 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
![\[(x + 1)^3 - x^3 = 181^2 \Rightarrow 3x(x + 1) = 181^2 - 1 \Rightarrow x(x + 1) = (60)(182).\]](http://latex.artofproblemsolving.com/7/d/9/7d9a2bdeba91e6922f38ddbfa31c60cd535feb57.png)
Since 
and 
, we might guess 
. Through this method or others, we find that and the sum of the two integers is 
.
