## Search found 73 matches

- Mon Feb 19, 2018 9:49 pm
- Forum: National Math Olympiad (BdMO)
- Topic: BdMO 2018 News
- Replies:
**4** - Views:
**681**

### Re: BdMO 2018 News

Neither BdMO nor APMO has a specific syllabus like school/college syllabus. And before the registration, add will be published on Prothom Alo.

- Mon Feb 19, 2018 9:12 pm
- Forum: National Math Olympiad (BdMO)
- Topic: BdMO 2018 News
- Replies:
**4** - Views:
**681**

### Re: BdMO 2018 News

Nothing to do. We need to attend both BdMO and APMO. So good luck!

- Mon Feb 19, 2018 8:47 pm
- Forum: National Math Olympiad (BdMO)
- Topic: BdMO 2018 News
- Replies:
**4** - Views:
**681**

### BdMO 2018 News

Dear participants of BdMO 2018 national round, the program is going to be held on 9-10 March, 2018 at the same location. Get more information here:

http://matholympiad.org.bd/

http://matholympiad.org.bd/

- Sun Feb 18, 2018 11:00 am
- Forum: Number Theory
- Topic: 13 -11
- Replies:
**6** - Views:
**3922**

### Re: 13 -11

**I missed the middle stamp!**

However a simple change will make the solution correct.

- Sun Feb 18, 2018 10:08 am
- Forum: Number Theory
- Topic: 13 -11
- Replies:
**6** - Views:
**3922**

### Re: 13 -11

Easy to prove by Modular Arithmetic $11|13^N-1$ We write: $13^N-1 \equiv 0 \Rightarrow 13^N \equiv 1 \Rightarrow (11+2)^N \equiv 1$ mod $(11)$ We can write the expression $(11+2)^N=11^N+{{N} \choose {1}} 11^{N-1}.2+{{N} \choose {2}} 11^{N-2}.2^2+...+{{N} \choose {N-1}} 11.2^{N-1}+2^N$ Here, all term...

- Fri Feb 16, 2018 7:45 pm
- Forum: Secondary Level
- Topic: secondary regional 2017
- Replies:
**9** - Views:
**1575**

### Re: secondary regional 2017

You didn't write that in the last question.

- Fri Feb 16, 2018 7:42 pm
- Forum: National Math Olympiad (BdMO)
- Topic: BdMO National Junior 2016/1
- Replies:
**7** - Views:
**1448**

### Re: BdMO National Junior 2016/1

We can write the expression in this form: $1+......(1+2(1+2(1+2)))...)$ where the number of brackets is $2016$. [We will call "Brackets" a pair of starting and ending brackets] If we take the whole expression under brackets we will get the number $2017$. If we encounter the first brackets, the resul...

- Fri Feb 16, 2018 6:56 pm
- Forum: Secondary Level
- Topic: secondary regional 2017
- Replies:
**9** - Views:
**1575**

### Re: secondary regional 2017

$34x+51y=6z \Rightarrow 34x=51y-6z=0$ $34x$ is even for all values of $x$. $6z$ is even too. Then, $51y$ must be even , $y=2$ then. We can write: $6z-34x=102$ [Applying the value of $y$] $ \Rightarrow 3z-17x=51$ $3$ divides RS . So, $3$ must divide $(3z-51)$. $3$ divides $3z$. Then $17x$. $3$ doesn'...

- Thu Feb 15, 2018 11:22 pm
- Forum: Secondary Level
- Topic: secondary regional 2017
- Replies:
**9** - Views:
**1575**

### Re: secondary regional 2017

Another clear-cut solution $34x+51y=6z \Rightarrow 34x+51y-6z=0$ Here, $34x$ and, $6z$ are even . So, $34x-6z$ is even . As LS is even $(0)$, $51y$ must be even . $51$ is not even . So, $y$ is obviously even and, $y=2$ We can write: $6z-34x=51 \times 2 \Rightarrow 6z-34x=102 \Rightarrow 3z-17x=51$ ...

- Thu Feb 15, 2018 4:16 pm
- Forum: National Math Olympiad (BdMO)
- Topic: BdMO National Primary 2016/2
- Replies:
**2** - Views:
**627**

### Re: BdMO National Primary 2016/2

$(11+11) \times (12+12) \times (13+13) \times...\times (18+18) \times (19+19)=22 \times 24 \times 26 \times...\times 38$

Here, the

As $x=30 \times y$

So, $10|x$

Remainder=$0$

Here, the

**fifth**term will be: $(15+15)=30$ which is divisible by $10$As $x=30 \times y$

So, $10|x$

Remainder=$0$

**[Can't simplify more for primary ]**