## Search found 73 matches

- Sat Mar 03, 2018 11:30 pm
- Forum: Secondary Level
- Topic: Greatest Positive Integer $x$
- Replies:
**7** - Views:
**5203**

### Greatest Positive Integer $x$

Find the greatest positive integer $x$ such that $23^{6+x}$ divides $2000!$

- Sat Mar 03, 2018 5:44 pm
- Forum: Secondary Level
- Topic: Right Triangle Geometry
- Replies:
**7** - Views:
**8707**

### Re: Right Triangle Geometry

So many solutions :!: Applying Sine law in $\triangle CDM$, we get: $\frac {CM}{sin \angle CDM}=\frac{DM}{sin \angle DCM} \Rightarrow \frac{CM}{\frac {1}{2}}=\frac{DM}{sin \angle DCM} \Rightarrow 2CM=\frac{DM}{sin \angle DCM} \Rightarrow DM=2CM \times sin\angle DCM$ ... (1) $M$ is the midpoint of t...

- Sat Mar 03, 2018 8:44 am
- Forum: Secondary Level
- Topic: Right Triangle Geometry
- Replies:
**7** - Views:
**8707**

### Re: Right Triangle Geometry

Perfect rational diagram it is

- Fri Mar 02, 2018 11:05 pm
- Forum: Secondary Level
- Topic: Boxes and Cards Problem
- Replies:
**2** - Views:
**3119**

### Boxes and Cards Problem

There are $ 7$ boxes arranged in a row and numbered $1$ through $7$. You have a stack of $2015$ cards, which you place one by one in the boxes. The first card is placed in box $\fbox {1}$ , the second in box $\fbox{2}$ , and so forth up to the seventh card which is placed in box $\fbox{7}$ . You the...

- Fri Mar 02, 2018 7:49 pm
- Forum: Secondary Level
- Topic: Right Triangle Geometry
- Replies:
**7** - Views:
**8707**

### Right Triangle Geometry

A famous and easy geometry problem: Let triangle $\triangle{ABC}$ have a right angle at $C$, and let $M$ be the midpoint of the hypotenuse $AB$. Choose a point $D$ on line $BC$ so that angle $\angle{CDM}$ measures $30$ degrees. Prove that the segments $AC$ and $MD$ have equal lengths. [Sorry if it ...

- Sat Feb 24, 2018 11:49 pm
- Forum: National Math Olympiad (BdMO)
- Topic: BdMO National Secondary 2006/10
- Replies:
**5** - Views:
**2161**

### Re: BdMO National Secondary 2006/10

Complicated. No $A$ found in the diagram.

- Sat Feb 24, 2018 11:43 pm
- Forum: National Math Olympiad (BdMO)
- Topic: BdMO National Higher Secondary 2014/9
- Replies:
**7** - Views:
**2793**

### Re: BdMO National Higher Secondary 2014/9

It is easy to say that finding the number of ways means finding the number of strings . If we denote the upper move with $U$ and the right move with $R$, we have $nU$ and $nR$ to create the dominated strings . At first, we assume there is no condition of mines . Then we have $2n$ places to arrange t...

- Tue Feb 20, 2018 10:59 am
- Forum: Divisional Math Olympiad
- Topic: Mymensingh Region Higher Secondary 2017/10
- Replies:
**2** - Views:
**1352**

### Re: Mymensingh Region Higher Secondary 2017/10

Think the case that an

Now, count for $4$ moves first, and then for last $2$ moves.

**Up**move will torpedo a**Down**move whence, a**Right**move will torpedo a**Left**move.Now, count for $4$ moves first, and then for last $2$ moves.

- Tue Feb 20, 2018 9:39 am
- Forum: National Math Olympiad (BdMO)
- Topic: BdMO 2017 National Round Secondary 7
- Replies:
**14** - Views:
**5412**

### Re: BdMO 2017 National Round Secondary 7

If the first case is true, you can't use two colors in $20$ pictures. The answer should be $9$

- Tue Feb 20, 2018 8:08 am
- Forum: National Math Olympiad (BdMO)
- Topic: BdMO 2017 National Round Secondary 7
- Replies:
**14** - Views:
**5412**

### Re: BdMO 2017 National Round Secondary 7

There is a common color in every $20$ pictures. My confusion is here. Does it mean just one common color in every 20 pictures or at least one among 20 ? If the first case is true, your solution is wrong. Because in $80$ pictures, you will get $2$ (B,G) common color. The answer should be $9$ If the ...