He said $\sum_{k\geq 1} \left \lfloor \frac{n}{p^k} \right \rfloor \ge p^a$ or if $p^x||n!$ then $x \geq a$, hope it clear things up.sourav das wrote:Actually, $\sum_{k\geq 1} \left \lfloor \frac{n}{p^k} \right \rfloor = $ The highest power that divide $n!$. So if thought $P^a||m$ and $P^a||n!$ then you have proved that $a \geq P^a$. Actually it doesn't make any sense to me. May be i couldn't get your solution correctly or you just make a bug. Please explain your solution to us more briefly.Phlembac Adib Hasan wrote: Now it's enough to show that $\sum_{k\geq 1} \left \lfloor \frac{n}{p^k} \right \rfloor \ge p^a$
Problem Set 3
Please read Forum Guide and Rules before you post.
Use $L^AT_EX$, It makes our work a lot easier!
Nur Muhammad Shafiullah | Mahi
Use $L^AT_EX$, It makes our work a lot easier!
Nur Muhammad Shafiullah | Mahi
- Phlembac Adib Hasan
- Posts:1016
- Joined:Tue Nov 22, 2011 7:49 pm
- Location:127.0.0.1
- Contact:
Re: Problem Set 3
Thank you.I missed the case.Now the proof has been edited.sourav via wrote:What if $x=2k$ where $(2,k)=1$ , $2|2$, $k|a$. Are you considering $x$ a prime?I didn't get your solution...Phlembac Adib Hasan wrote:As $2$ and $a$ are co-primes, $x|2$ or $x|a$
Welcome to BdMO Online Forum. Check out Forum Guides & Rules
- Phlembac Adib Hasan
- Posts:1016
- Joined:Tue Nov 22, 2011 7:49 pm
- Location:127.0.0.1
- Contact:
Re: Problem Set 3
If $(a,b)=1$ then it's not always true $(2a,\phi (b) )=2$ because $\phi (b)$ may have such prime factors that $b$ doesn't have.Nadim Ul Abrar wrote:No 8:
Welcome to BdMO Online Forum. Check out Forum Guides & Rules
-
- Posts:461
- Joined:Wed Dec 15, 2010 10:05 am
- Location:Dhaka
- Contact:
Re: Problem Set 3
I meant to actually he has to proof $\sum_{k\geq 1} \left \lfloor \frac{n}{p^k} \right \rfloor=a \ge x$ where he proved that $a\geq P^x$*Mahi* wrote:He said $\sum_{k\geq 1} \left \lfloor \frac{n}{p^k} \right \rfloor \ge p^a$ or if $p^x||n!$ then $x \geq a$, hope it clear things up.sourav das wrote:Actually, $\sum_{k\geq 1} \left \lfloor \frac{n}{p^k} \right \rfloor = $ The highest power that divide $n!$. So if thought $P^a||m$ and $P^a||n!$ then you have proved that $a \geq P^a$. Actually it doesn't make any sense to me. May be i couldn't get your solution correctly or you just make a bug. Please explain your solution to us more briefly.Phlembac Adib Hasan wrote: Now it's enough to show that $\sum_{k\geq 1} \left \lfloor \frac{n}{p^k} \right \rfloor \ge p^a$
You spin my head right round right round,
When you go down, when you go down down......(-$from$ "$THE$ $UGLY$ $TRUTH$" )
When you go down, when you go down down......(-$from$ "$THE$ $UGLY$ $TRUTH$" )
- Nadim Ul Abrar
- Posts:244
- Joined:Sat May 07, 2011 12:36 pm
- Location:B.A.R.D , kotbari , Comilla
Re: Problem Set 3
Still there are mistakes. You know that $b$ does not share any common factor with $a$ but $\varphi(b)$ may share, because there are some $p_i-1$'s in $\varphi(b)$ who may contain a prime factor of $a$.
One one thing is neutral in the universe, that is $0$.