Question

令 $A$ 为 $2025$ 的正整数因数集合。令 $B$ 为从 $A$ 中随机选取的一个子集。 $B$ 是非空集合且其元素的最小公倍数为 $2025$ 的概率为 $\frac{m}{n}$ ，其中 $m$ 和 $n$ 是互质正整数。求 $m + n$ 。

Let $A$ be the set of positive integer divisors of $2025.$ Let $B$ be a randomly selected subset of $A.$ The probability that $B$ is a nonempty set with the property that the least common multiple of its elements is $2025$ is $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$

答案

答案：237

知识点：最小公倍数容斥原理子集

难度评级：2510

解答：

因为 $2025 = 3^4 \cdot 5^2$ ，集合 $A$ 有 $5 \cdot 3 = 15$ 个元素，所以共有 $2^{15}$ 个子集。一个子集的最小公倍数为 $2025$ 当且仅当它至少含有一个可被 $3^4 = 81$ 整除的因数，并且至少含有一个可被 $5^2 = 25$ 整除的因数（这样的子集自动非空）。不被 $81$ 整除的因数有 $12$ 个，不被 $25$ 整除的因数有 $10$ 个，二者都不整除的有 $8$ 个。

由容斥，符合条件的子集数为因为 $27904 = 2^8 \cdot 109$ ，概率为 $\frac{27904}{32768} = \frac{109}{128}$ ，所以 $m + n = 109 + 128 = 237$ 。 $\begin{gathered} 2^{15} - 2^{12} - 2^{10} + 2^8 \\ = 32768 - 4096 - 1024 + 256 \\ = 27904. \end{gathered}$

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 $m + n = 109 + 128 = 237.$

2025 AIME II 第 7 题

答案：237

难度评级：2510

解答：

其他年份的第 7 题