2025 AIME II 第 7 题

$2025 = 3^4 \cdot 5^2$ has $15$ divisors; the lcm is $2025$ exactly when the subset contains a multiple of $81$ and a multiple of $25$

Use inclusion-exclusion: $2^{12}$ subsets avoid all three multiples of $81,$ and $2^{10}$ avoid all five multiples of $25$

Since $2025 = 3^4 \cdot 5^2,$ the set $A$ has $5 \cdot 3 = 15$ elements, and there are $2^{15}$ subsets. A subset has least common multiple $2025$ exactly when it contains at least one divisor divisible by $3^4 = 81$ and at least one divisible by $5^2 = 25$ (such a subset is automatically nonempty). There are $12$ divisors not divisible by $81,$ $10$ not divisible by $25,$ and $8$ divisible by neither.

By inclusion-exclusion, the number of good subsets is $$\begin{gathered} 2^{15} - 2^{12} - 2^{10} + 2^8 \\ = 32768 - 4096 - 1024 + 256 \\ = 27904. \end{gathered}$$ Since $27904 = 2^8 \cdot 109,$ the probability is $\frac{27904}{32768} = \frac{109}{128},$ and $m + n = 109 + 128 = 237.$