Question

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 $m + n.$

Answer

Solution:

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 $2^{15} - 2^{12} - 2^{10} + 2^8 = 32768 - 4096 - 1024 + 256 = 27904.$ Since $27904 = 2^8 \cdot 109,$ the probability is $\frac{27904}{32768} = \frac{109}{128},$ and $m + n = 109 + 128 = 237.$

2025 AIME II Problem 7

Solution:

Problem 7 in Other Years