Difference between revisions of "2025 AIME II Problems/Problem 7"
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Let <math>A</math> be the set of positive integer divisors of <math>2025</math>. Let <math>B</math> be a randomly selected subset of <math>A</math>. The probability that <math>B</math> is a nonempty set with the property that the least common multiple of its element is <math>2025</math> is <math>\frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>. | Let <math>A</math> be the set of positive integer divisors of <math>2025</math>. Let <math>B</math> be a randomly selected subset of <math>A</math>. The probability that <math>B</math> is a nonempty set with the property that the least common multiple of its element is <math>2025</math> is <math>\frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>. | ||
| − | == Solution == | + | == 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 | ||
Revision as of 22:16, 13 February 2025
Problem
Let 
be the set of positive integer divisors of 
. Let 
be a randomly selected subset of . The probability that is a nonempty set with the property that the least common multiple of its element is is 
, where 
and 
are relatively prime positive integers. Find 
.
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
