## Search found 86 matches

- Mon Jan 23, 2017 8:18 am
- Forum: Divisional Math Olympiad
- Topic: Divisional MO 2015
- Replies:
**4** - Views:
**886**

### Re: Divisional MO 2015

See, $9800 = 2^3 \times 5^2 \times 7^2$ Let the two numbers will be $p = 2^a \times 5^b \times 7^c$ and $q = 2^x \times 5^y \times 7^z$. Now consider the pairs of $(a,x)$ which are possible: $(0,3), (1,3),(2,3),(3,3), (3,0), (3,1), (3,2),(1,2),(2,1)$ , numbers of pairs is $9$. Again, similarly the p...

- Thu Jan 19, 2017 6:54 pm
- Forum: Secondary Level
- Topic: Dhaka Regional '16 P8
- Replies:
**3** - Views:
**960**

### Re: Dhaka Regional '16 P8

I have to solve it like this!?

- Thu Jan 19, 2017 2:55 am
- Forum: Secondary Level
- Topic: Dhaka Regional '16 P8
- Replies:
**3** - Views:
**960**

### Dhaka Regional '16 P8

How many eight digit number can be formed by using the digits $1, 2, 3, 4, 5, 6, 7, 8$ so that each number has $6$ digits in such place where that digit is less than the next digit? Example: In number $2314$; $2, 1$ are two digits such that each of them is less than the next digit. Translated into B...

- Thu Jan 19, 2017 2:13 am
- Forum: National Math Olympiad (BdMO)
- Topic: National BDMO 2016 : Junior 8
- Replies:
**4** - Views:
**1265**

### Re: National BDMO 2016 : Junior 8

Solution: Take $X,Y$ reflections of $B$ about $P,Q$ respectively and $Z$ reflection of $C$ about $P$.Now see that $\triangle ZYC$ is equilateral triangle, so $ZY=ZC=ZD$, $Z$ is circumcenter of $\triangle CXY$, thus $\angle CXY=30^\circ$, but $PQ$ is midline of $\triangle CXY$, so $PQ\parallel XY$, t...

- Thu Jan 19, 2017 2:00 am
- Forum: Site Support
- Topic: How to prepare and study for BdMO in Secondary Group
- Replies:
**3** - Views:
**1849**

### Re: How to prepare and study for BdMO in Secondary Group

Adib Hasan vaia wrote a note about "Gonit Olympiad Preparation", you can follow those instructions that will be helpful for you I think.

- Sun Dec 25, 2016 11:35 am
- Forum: Number Theory
- Topic: USAMO 2015 P5
- Replies:
**2** - Views:
**736**

### USAMO 2015 P5

Let $a, b, c, d$ be distinct positive integers such that $a^4 + b^4 = c^4 + d^4 = e^5 $. Show that $ ac + bd $ is a composite number.

Can't solve it properly . Any help will be appreciated!

Thanks You!

Can't solve it properly . Any help will be appreciated!

Thanks You!

- Fri Dec 23, 2016 9:18 pm
- Forum: Algebra
- Topic: Blended with equations
- Replies:
**1** - Views:
**436**

### Re: Blended with equations

Given that, $ab+bc =130 ... ... ... ... ...$ (i) $bc+ca =168 ... ... ... ... ...$ (ii) $ca+ab =228 ... ... ... ... ...$ (iii) Sum this three equation and you will get, $ 2 (ab + bc + ca ) = 586 $ $\Rightarrow ab + bc + ca = 263 ... ... ... ... ...$ (iv) From (i) and (iv) we get, $ 130 + ca = 263 $ $...

- Tue Dec 20, 2016 6:19 pm
- Forum: Site Support
- Topic: Problem! Help anybody!
- Replies:
**4** - Views:
**1252**

### Re: Problem! Help anybody!

Yes, it is working now.

**THANK YOU**, Vaia.- Tue Dec 20, 2016 2:47 pm
- Forum: Combinatorics
- Topic: Boy and Girls!
- Replies:
**0** - Views:
**639**

### Boy and Girls!

There are $n$ boys and $n+1$ girls standing in a straight line in some order (from left to right). A group of girls is a sequence of at least two neighboring girls with no girl immediately to the left and no girl immediately to the right of it. In a move, we take the leftmost group of girls and make...

- Tue Dec 20, 2016 11:21 am
- Forum: Number Theory
- Topic: Pretty Diophantine Equation
- Replies:
**2** - Views:
**909**

### Pretty Diophantine Equation

Find all pairs of integers $ (x,y)$, such that

\[ x^2 - 2009y + 2y^2 = 0

\]

\[ x^2 - 2009y + 2y^2 = 0

\]