+35 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 $f(n) > 15$?
THANKS
+118680 Hi EggRick,
How about presenting your questions in the best possible way.
Nobody wants to look at all those irrelevant dollar signs.
+35 Sorry!!!
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?
THANKS
The corrected response for your question is:
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. We are given that n is a multiple of 15. Since 15 = 3 x 5, we know that any n 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 pi is the power of the ith prime factor in k!.
Now, consider an integer n that is a multiple of 15. The smallest possible value of k such that k! is divisible by n 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 >= 1, and all other prime factors have exponent 0.
We are given that f(n) > 15, so we can write:
2a * 3b * 5c * ... > 15
Simplifying this inequality, we get:
2^(a-3) * 3^(b-2) * 5^(c-1) * ... > 1
Since a, b, and c are integers, we must have:
a >= 3, b >= 2, c >= 1
Therefore, the smallest possible value of n is:
n = 2^3 * 3^2 * 5 = 360
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.
