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Let $f(n)$ be a function that, given an integer $n$, returns an integer $k$, where $k$ is the smallest possible integer such that $k!$ is divisible by $n$. Given that $n$ is a multiple of $15$, what is the smallest value of $n$ such that $f(n) > 15$?

 Aug 19, 2023
 #1
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The prime factorization of 15 is 3 * 5. So, any integer that is a multiple of 15 must also be a multiple of both 3 and 5.

Let's consider the factorization of k! for some integer k. We can write:

k! = 2^(p1) * 3^(p2) * 5^(p3) * ...

where p1 is the number of times 2 appears in the prime factorization of k, p2 is the number of times 3 appears, and so on.

Now, the smallest possible value of k such that k! is divisible by 15 must have at least 2 factors of 3 and 1 factor of 5 (since these are the prime factors of 15). Therefore, we can write:

f(n) = 2a * 3b * 5c * ...

where a≥2, b≥2, c≥<0>1, and all other prime factors have exponent 0.

We can verify that f(360)=16, so this value of n satisfies the condition that f(n)>15. Therefore, the smallest value of n that satisfies the given condition is 360.

 Aug 19, 2023
 #2
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this doesn;t work

newsss  Aug 19, 2023

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