Name you'd like to be called: Epshita

Course you want to learn: Functional Equations and Number Theory Problem solving.

Preferred methods of communication (Forum, Messenger, Telegram, etc.):Telegram

Do you want to take lessons through PMs or Public?: Public

## Search found 37 matches

- Tue Mar 28, 2017 11:14 pm
- Forum: National Math Camp
- Topic: The Gonit IshChool Project - Beta
- Replies:
**28** - Views:
**31359**

- Tue Mar 28, 2017 8:32 pm
- Forum: Social Lounge
- Topic: BDMO Forum Mafia #1
- Replies:
**52** - Views:
**22533**

### Re: BDMO Forum Mafia #1

**VOTE: (thamimzahin)**

Reason: Same as ahmedittihad. And also, he has no use living.

- Mon Mar 27, 2017 9:06 pm
- Forum: Site Support
- Topic: Help
- Replies:
**1** - Views:
**7014**

### Re: Help

Well, your ipad is shit (forgive my rudeness, bdmo forum admins and mods). Get a better device.

- Mon Mar 27, 2017 6:21 pm
- Forum: Social Lounge
- Topic: BDMO Forum Mafia #1
- Replies:
**52** - Views:
**22533**

### Re: BDMO Forum Mafia #1

Oka, I am in.

- Fri Mar 24, 2017 9:24 pm
- Forum: Social Lounge
- Topic: BDMO Forum Mafia
- Replies:
**5** - Views:
**2151**

### Re: BDMO Forum Mafia

By pm-ing?

- Fri Mar 24, 2017 2:57 am
- Forum: Junior Level
- Topic: Beginner's Marathon
- Replies:
**68** - Views:
**18916**

### Re: Beginner's Marathon

P3. Several stones are placed on an infinite (in both directions) strip of squares. As long as there are at least two stones on a single square, you may pick up two such stones, then move one to the preceding square and one to the following square. Is it possible to return to the starting configurat...

- Thu Mar 23, 2017 12:18 am
- Forum: Junior Level
- Topic: Beginner's Marathon
- Replies:
**68** - Views:
**18916**

### Re: Beginner's Marathon

It's the max number of edges in a disconnected graph + 1.

So suppoose we have a maximal disconnected graph. Then each connected component must be a clique.

So answer = K_2003 + isolated vertex.

So suppoose we have a maximal disconnected graph. Then each connected component must be a clique.

So answer = K_2003 + isolated vertex.

- Sun Sep 06, 2015 10:54 am
- Forum: National Math Camp
- Topic: ONTC Final Exam
- Replies:
**34** - Views:
**16358**

### Re: ONTC Final Exam

When will you post the solutions ? And I was thinking of new ways to solve p2 . Can it be done by congruence ?

- Fri Sep 04, 2015 11:09 am
- Forum: National Math Camp
- Topic: ONTC Final Exam
- Replies:
**34** - Views:
**16358**

### Re: ONTC Final Exam

Masum bai , I know this is not a proof . I was talking about my proof that I haven't posted yet . And thank you . My solution consists of the hint you gave . I was thinking of this was correct or not .

- Thu Sep 03, 2015 3:41 pm
- Forum: National Math Camp
- Topic: ONTC Final Exam
- Replies:
**34** - Views:
**16358**

### Re: ONTC Final Exam

In p2 , 2^m-1 according to m makes a sequence - if m = 1 , then 2^m-1 = 1 . For m as 2 , 2^m-1 = 3 . For m as 3 , 2^m-1 = 7 . Let 2^m-1 = X . The sequence - Xn = 2*X(n-1) + 1 where X1 = 1 . In the whole sequence only when m = 1 , X divides n^2+1 . For ex : n^2+1 = 1, 2 (mod 3) etc . My proof is poor...