Difference between revisions of "2025 AIME II Problems/Problem 7"

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~Mitsuihisashi14
 
~Mitsuihisashi14

Revision as of 22:16, 13 February 2025

Problem

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 element is $2025$ is $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

Solution 1

We split into different conditions:

Note that the set needs to satisfy with all of the elements' lcm = 2025, we need to ensure that the set has at least 1 number that is a multiple of 3^4 and a number that is a multiple of 5^2.

Multiples of 3^4: 81, 405, 2025 Multiples of 5^2: 25, 75, 225, 675, 2025

If the set B contains 2025, then all of the rest 14 factors is no longer important. The valid cases are 2^14.

If the set B doesn't contain 2025, but contains 405, we just need another multiple of 5^2. It could be 1 of them, 2 of them, 3 of them or 4 of them, which has 2^4 - 1 = 15 cases. Excluding 2025, 405, 25, 75, 225, 675, the rest 9 numbers could appear or not appear. Therefore, this case has a valid case of 15 * 2^9.

If set B doesn't contain 2025 nor 405, it must contain 81. It also needs to contain at least 1 of the multiples from 5^2, where it would be 15 * 2^8.

The total valid cases are 2^14 + 15 * (2^9 + 2^8), and the total cases are 2^15. The answer is 2^8 * (64 + 30 + 15) / 2^8 * 2^7 = 109/128. Desired answer: 109 + 128 = 237

(Feel free to edit any latex or format problems)

~Mitsuihisashi14