Prove that for any integer $K$, There exists a natural number $N$ so that in the prime factorization of $N!$ we can find at least $K$ prime numbers which their powers are exactly $1$.
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1$\begingroup$ Welcome and remember! The use of $\LaTeX$ syntax is highly recommended! $\endgroup$– Sangchul LeeCommented May 6, 2013 at 15:11
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$\begingroup$ Okay! I will try to use it for my next questions! $\endgroup$– CODECommented May 6, 2013 at 15:30
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2 Answers
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The primes in $(N/2,N]$ appear with exponent 1, so you need to show that there are arbitrarily many primes in such an interval. You could use Ramanujan's proof of Bertrand's postulate, for example.
If you have access to sledgehammers like the Prime Number Theorem that would suffice as well.
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Hint Prove that there exists some $n$ so that $\pi(2n)-\pi(n)=K$, where $\pi$ is the prime counting function. Then you can show that $(2n)!$ has the desired result.
To prove this hint, try combining the Prime Number Theorem with the fact that $f(n)=\pi(2n)-\pi(n)$ satisfies
$$f(n+1)-f(n) \in \{ 0, 1, -1 \} \,;\, \forall n\ geq 3 \,.$$
