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

Revision as of 03:05, 14 February 2025

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

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

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

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

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

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

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

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

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

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

So n=-1, 1, 11 and 37

Since n is positive, so n=1, 11 and 37

1+11+37= $\framebox{49}$ is the correct answer

～Tonyttian [1]

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 ～Tonyttian [https://artofproblemsolving.com/wiki/index.php/User:Tonyttian]
+==See also==
+{{AIME box|year=2025|num-b=1|num-a=3|n=II}}
+{{MAA Notice}}

2025 AIME II (Problems • Answer Key • Resources)
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All AIME Problems and Solutions