Search found 327 matches
- Tue Aug 28, 2012 12:30 am
- Forum: Combinatorics
- Topic: Quite hard !!! (may be)
- Replies: 4
- Views: 3683
Re: Quite hard !!! (may be)
$n!e=n!(1+\frac{1}{2!}+\frac{1}{3!}+...)=(\displaystyle\sum_{k=0}^{n}\frac{n!}{(n-k)!})+\frac{n!}{(n+1)!}+\frac{n!}{(n+2)!}+...$ All we have got to do next is to show that $\frac{n!}{(n+1)!}+\frac{n!}{(n+2)!}+...=\frac{1}{n+1}+\frac{1}{n+2}...<1$. Can we do that? may be here is a typo (i'm not sure...
- Sun Aug 26, 2012 11:15 pm
- Forum: Combinatorics
- Topic: Quite hard !!! (may be)
- Replies: 4
- Views: 3683
Quite hard !!! (may be)
Show that for $n \in \mathbb{N}$,
$\displaystyle\sum_ {k = 0}^{n} n \mathrm{P}_k = \left\lfloor{n!e}\right\rfloor $
Source: The Collage Math J.20(1989),260
$\displaystyle\sum_ {k = 0}^{n} n \mathrm{P}_k = \left\lfloor{n!e}\right\rfloor $
Source: The Collage Math J.20(1989),260
- Thu Aug 09, 2012 12:57 pm
- Forum: Asian Pacific Math Olympiad (APMO)
- Topic: APMO 1990 (Inequality with Combi)
- Replies: 2
- Views: 3592
Re: APMO 1990 (Inequality with Combi)
Well here is my solution Let prove a quick lemma Lemma:For finite n element set, the product of $a_{1},a_{2}.......,a_{n}$ taken $k$ at a time is $[a_{1}a_{2}.......a_{n}]^{r}$ where $r_{1}=\displaystyle \binom{n-1}{k-1}$ because here every positive integer repeated for $r_{1}$ times. Now back to ma...
- Sun Aug 05, 2012 11:57 am
- Forum: International Mathematical Olympiad (IMO)
- Topic: IMO 2012: Day 1 Problem 2
- Replies: 12
- Views: 16185
Re: IMO 2012: Day 1 Problem 2
Hey,nice problem. I also use the similar technique like Tahmid.
- Mon Jun 11, 2012 1:03 am
- Forum: Secondary Level
- Topic: Even,You are odd looking!!!!!!!!!
- Replies: 8
- Views: 5800
Re: Even,You are odd looking!!!!!!!!!
OOps,not just this part,I think ur solution is totally wrong.Because,After summing, we get,\[10^{16}(x_{1}+x_{17})+10^{15}(x_{2}+x_{16})+10^{14}(x_{3}+x_{15})+...\] Look that,\[(x_{1}+x_{17}),(x_{2}+x_{16}),(x_{3}+x_{15})+...\] can be greater than $9$. So they are not digits(1,2,......9,0) so 1 can...
- Sat Jun 09, 2012 6:43 pm
- Forum: Secondary Level
- Topic: Even,You are odd looking!!!!!!!!!
- Replies: 8
- Views: 5800
Re: Even,You are odd looking!!!!!!!!!
Hey,No one interested :? ??? That's very easy one so may be no one want to make a reply. Anyway there is a bug in your problem. You said it is a 17 digit number but your $N$ and $M$ representation says they are 19 digit number ;) ok it's not a matter you can also solve it. here is the solve but now...
- Fri Jun 08, 2012 2:30 pm
- Forum: Asian Pacific Math Olympiad (APMO)
- Topic: APMO-2012 results
- Replies: 6
- Views: 5392
Re: APMO-2012 results
Finally the official results of APMO are published http://www.mmjp.or.jp/competitions/APMO/ congratulations to Bappa vai for getting Silver;Mugdho vai and Sourav vai for getting Bronze and also Adib and Choton vai for getting honourable mentions. and the overall position of Bangladesh is 19th.Thank...
- Wed Jun 06, 2012 11:06 pm
- Forum: Junior Level
- Topic: No repeated digits: 100-999
- Replies: 8
- Views: 6239
Re: No repeated digits: 100-999
Amar ager ans ta vul chilo....tarahura kore koray vul hoiche..keu notice kore nai. :lol: দেখছিলাম কিন্তু রিপ্লাই দেয়ার টাইম পাই নাই। যাইহোক তোমার মেথড টা ঠিকাছে। এটাকে জেনেরালাইজ কর আরও অনেক মজার জিনিস পাবা। আর এইটা নিয়ে আরও জোস প্রব্লম আছে। আমি কোন জায়গায় জানি দিসিলাম। এই ফোরামেই। খুজে দেখ। জো...
- Fri Jun 01, 2012 11:04 pm
- Forum: College / University Level
- Topic: Dangerous integretion :P
- Replies: 5
- Views: 10424
Dangerous integretion :P
Please, any one show me the procedure to get the integral of $sin x^{2} dx$ ????
- Sun May 27, 2012 8:40 pm
- Forum: Social Lounge
- Topic: অভিনন্দন !
- Replies: 12
- Views: 9972
Re: অভিনন্দন !
#incude<অভিনন্দন.h> int main() { while(scanf==1) { printf("শুভেচ্ছা"); } if (মেডেল==Bronze) { printf("অভিনন্দন"); } if (মেডেল==Silver) { printf("উরাধুরা অভিনন্দন"); } if (মেডেল==Gold) { printf("তোরে আর কি অভিনন্দন দিমু রে পাগলা। তুই তো কোপাইয়া দিসস।"); } return ভালোবাসা; } :lol: :lol: :lol: :lol: :...