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Difference between revisions of "2025 AIME II Problems/Problem 2"

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Find the sum of all positive integers <math>n</math> such that <math>n + 2</math> divides the product <math>3(n + 3)(n^2 + 9)</math>.
 
Find the sum of all positive integers <math>n</math> such that <math>n + 2</math> divides the product <math>3(n + 3)(n^2 + 9)</math>.
  
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==Solution 1==
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<math>\frac{3(n+3)(n^{2}+9) }{n+2} \in Z</math>
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<math>\Rightarrow \frac{3(n+2+1)(n^{2}+9) }{n+2} \in Z</math>
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<math>\Rightarrow \frac{3(n+2)(n^{2}+9) +3(n^{2}+9)}{n+2} \in Z</math>
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<math>\Rightarrow 3(n^{2}+9)+\frac{3(n^{2}+9)}{n+2} \in Z</math>
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<math>\Rightarrow \frac{3(n^{2}-4+13)}{n+2} \in Z</math>
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<math>\Rightarrow \frac{3(n+2)(n-2)+39}{n+2} \in Z</math>
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<math>\Rightarrow 3(n-2)+\frac{39}{n+2} \in Z</math>
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<math>\Rightarrow \frac{39}{n+2} \in Z</math>
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Since <math>n + 2</math> is positive, the positive factors of <math>39</math> are <math>1</math>, <math>3</math>, <math>13</math>, and <math>39</math>.
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Therefore, <math>n = -1</math>, <math>1</math>, <math>11</math> and <math>37</math>.
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Since <math>n</math> is positive, <math>n = 1</math>, <math>11</math> and <math>37</math>.
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<math>1 + 11 + 37 = \framebox{049}</math> is the correct answer
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～[https://artofproblemsolving.com/wiki/index.php/User:Tonyttian Tonyttian]
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~ Edited by [https://artofproblemsolving.com/wiki/index.php/User:Aoum aoum]
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==See also==
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{{AIME box|year=2025|num-b=1|num-a=3|n=II}}
  
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==Solution 1==
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{{MAA Notice}}
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[Please input a solution.]
 

Latest revision as of 13:05, 15 February 2025

Problem

Find the sum of all positive integers $n$ such that $n + 2$ divides the product $3(n + 3)(n^2 + 9)$.

Solution 1

$\frac{3(n+3)(n^{2}+9) }{n+2} \in Z$

$\Rightarrow \frac{3(n+2+1)(n^{2}+9) }{n+2} \in Z$

$\Rightarrow \frac{3(n+2)(n^{2}+9) +3(n^{2}+9)}{n+2} \in Z$

$\Rightarrow 3(n^{2}+9)+\frac{3(n^{2}+9)}{n+2} \in Z$

$\Rightarrow \frac{3(n^{2}-4+13)}{n+2} \in Z$

$\Rightarrow \frac{3(n+2)(n-2)+39}{n+2} \in Z$

$\Rightarrow 3(n-2)+\frac{39}{n+2} \in Z$

$\Rightarrow \frac{39}{n+2} \in Z$

Since $n + 2$ is positive, the positive factors of $39$ are $1$, $3$, $13$, and $39$.

Therefore, $n = -1$, $1$, $11$ and $37$.

Since $n$ is positive, $n = 1$, $11$ and $37$.

$1 + 11 + 37 = \framebox{049}$ is the correct answer

～Tonyttian

~ Edited by aoum

See also

2025 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 1 		Followed by
Problem 3
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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