Difference between revisions of "2025 AIME II Problems/Problem 2"
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<math>\Rightarrow \frac{3(n+2)(n-2)+39}{n+2} \in Z</math> | <math>\Rightarrow \frac{3(n+2)(n-2)+39}{n+2} \in Z</math> | ||
− | <math>\Rightarrow 3(n | + | <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 | Since n+2 is positive,the positive factor of 39 are 1, 3, 13 and 39 |
Revision as of 21:12, 13 February 2025
Problem
Find the sum of all positive integers such that
divides the product
.
Solution 1
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= is the correct answer
~Tonyttian [1]