Difference between revisions of "2025 AIME II Problems/Problem 7"
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| + | ==Solution 2== | ||
| + | We take the complement and use PIE. Suppose the LCM of the elements of the set is NOT <math>2025</math>. Since <math>2025=3^4 \cdot 5^2</math>, it must be that no element <math>x</math> in the subset satisfies <math>v_3(x)=4</math> OR no element <math>x</math> in the subset satisfies <math>v_5(x)=2</math> (in this case, <math>v_p(n)</math> gives the exponent of <math>p</math> in the prime factorization of <math>n</math>). For the first case, there are <math>4 \cdot 3 = 12</math> possible divisors that could be in our subset (<math>v_3(x)=0,1,2,3,v_5(x)=0,1,2</math> are possible), for a total count of <math>2^{12}</math> subsets. In the second case, there are <math>5 \cdot 2 = 10</math> possible divisors that could be in our subset, for a total count of <math>2^{10}</math> subsets. However, if both conditions are violated, then there are <math>4 \cdot 2 = 8</math> divisors that could be in our subset, for a total count of <math>2^8</math> subsets. Hence, by PIE, the number of subsets whose LCM is NOT <math>2025</math> is equal to <math>2^{12}+2^{10}-2^8</math>. The answer is then <cmath>1-\frac{2^{12}+2^{10}-2^8}{2^{5 \cdot 3}}=\frac{109}{128} \implies \boxed{237}.</cmath> | ||
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| + | ~cxsmi | ||
Revision as of 23:00, 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
Solution 2
We take the complement and use PIE. Suppose the LCM of the elements of the set is NOT . Since 
, it must be that no element 
in the subset satisfies 
OR no element in the subset satisfies 
(in this case, 
gives the exponent of 
in the prime factorization of ). For the first case, there are 
possible divisors that could be in our subset (
are possible), for a total count of 
subsets. In the second case, there are 
possible divisors that could be in our subset, for a total count of 
subsets. However, if both conditions are violated, then there are 
divisors that could be in our subset, for a total count of 
subsets. Hence, by PIE, the number of subsets whose LCM is NOT is equal to 
. The answer is then ![\[1-\frac{2^{12}+2^{10}-2^8}{2^{5 \cdot 3}}=\frac{109}{128} \implies \boxed{237}.\]](http://latex.artofproblemsolving.com/2/1/9/21978a63d8caa4d5b0d46d0c9e0b657e77c8dbcf.png)
~cxsmi
