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

Revision as of 21:03, 13 February 2025

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

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

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

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

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

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

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

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

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

So x=-1, 1, 11 and 37

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

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

～Tonyttian [1]

@@ Line 1: / Line 1: @@
 ==Problem==
 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>.
+==Solution 1==
+<math>\frac{3(n+3)(n^{2}+9) }{n+2} \in Z</math>
+<math>\Rightarrow \frac{3(n+2+1)(n^{2}+9) }{n+2} \in Z</math>
+<math>\Rightarrow \frac{3(n+2)(n^{2}+9) +3(n^{2}+9)}{n+2} \in Z</math>
+<math>\Rightarrow 3(n^{2}+9)+\frac{3(n^{2}+9)}{n+2} \in Z</math>
+<math>\Rightarrow \frac{3(n^{2}-4+13)}{n+2} \in Z</math>
+<math>\Rightarrow \frac{3(n+2)(n-2)+39}{n+2} \in Z</math>
+<math>\Rightarrow 3(n+2)+\frac{39}{n+2} \in Z</math>
+Since n+2 is positive,the positive factor of 39 are 1, 3, 13 and 39
+So x=-1, 1, 11 and 37
+Since x is positive, so x=1, 11 and 37
++11+37=  <math>\framebox{49}</math> is the correct answer
+～Tonyttian [https://artofproblemsolving.com/wiki/index.php/User:Tonyttian]