Too easy(Advanced)

[sakibtanvir]

I built this being inspired by Pritom vaia's problem.
Prove that, The number of prime numbers from 1 to n is less than $(2n/5)$ .

[Phlembac Adib Hasan]

I built this being inspired by Pritom vaia's problem.
Prove that, The number of prime numbers from 1 to n is less than $(2n/5)$ .

An advice : you should plug in smaller values before stating a problem.Because it may bring a limit.For example, this problem is continuously true only for $n\ge 21$.
Proof :

A super easy problem.I needed only two minutes to solve.Simple induction.It's known to all that every odd prime greater than $3$ is of the form either $6k+1$ or $6k-1$.
The base case is true for $n=21,...,25$.Now let $n=6k+1$ and follows the condition.It's nearest prime can be $6k+5$.So $\pi (n)$ remains unchanged till $n=6k+4$.But $\frac {2n}{5}$ increases.So the inequality remains valid.And for $n=6k+5$ , $\pi (n)$ can increase at most by $1$.But we assumed at first that $\pi(6k+1)<\frac{12k+2}{5}$.So it follows that \[\pi(6k+5)\leq\pi (6k+1)+1<\frac{12k+2}{5}+1=\frac {12k+7}{5}<\frac {12k+10}{5}=\frac {2(6k+5)}{5}\]Now I have to prove the condition is true for $n=6(k+1)+1$.Like the previous argument I can say \[\pi(6(k+1)+1)\leq \pi(6k+1)+2<\frac {12k+2}{5}+2\]\[=\frac {12k+12}{5}<\frac {12k+14}{5}=\frac {2[6(k+1)+1]}{5}\]Which completes the induction.

A generalization :My Proof shows that $\pi(n)<\frac {n}{3}$ is true for every $n\ge34$.

[nafistiham]

I built this being inspired by Pritom vaia's problem.
Prove that, The number of prime numbers from 1 to n is less than $(2n/5)$ .

An advice : you should plug in smaller values before stating a problem.Because it may bring a limit.For example, this problem is continuously true only for $n\ge 21$.
Proof :

A super easy problem.I needed only two minutes to solve.Simple induction.It's known to all that every odd prime greater than $3$ is of the form either $6k+1$ or $6k-1$.
The base case is true for $n=21,...,25$.Now let $n=6k+1$ and follows the condition.It's nearest prime can be $6k+5$.So $\pi (n)$ remains unchanged till $n=6k+4$.But $\frac {2n}{5}$ increases.So the inequality remains valid.And for $n=6k+5$ , $\pi (n)$ can increase at most by $1$.But we assumed at first that $\pi(6k+1)<\frac{12k+2}{5}$.So it follows that \[\pi(6k+5)\leq\pi (6k+1)+1<\frac{12k+2}{5}+1=\frac {12k+7}{5}<\frac {12k+10}{5}=\frac {2(6k+5)}{5}\]Now I have to prove the condition is true for $n=6(k+1)+1$.Like the previous argument I can say \[\pi(6(k+1)+1)\leq \pi(6k+1)+2<\frac {12k+2}{5}+2\]\[=\frac {12k+12}{5}<\frac {12k+14}{5}=\frac {2[6(k+1)+1]}{5}\]Which completes the induction.

A generalization :My Proof shows that $\pi(n)<\frac {n}{3}$ is true for every $n\ge34$.

why don't you share the general proof ?

[Phlembac Adib Hasan]

why don't you share the general proof ?

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