## Search found 13 matches

- Thu Jan 14, 2016 12:39 am
- Forum: Higher Secondary Level
- Topic: A problem of combinatorics
- Replies:
**9** - Views:
**11558**

### Re: A problem of combinatorics

A person can shake hands with others for maximum $35$ times. Suppose $A_1$ shook hands 35 times with $A_2$,$A_3$....$A_{36}$. Then $A_2$ can shake hands maximum $34$ times.As he has already shaken hands with $A_1$, suppose he hasn't shaken hands with $A_3$.Similarly $A_3$ hasn't shaken hands with $A...

- Mon Jan 04, 2016 11:23 pm
- Forum: International Mathematical Olympiad (IMO)
- Topic: IMO 1974/1
- Replies:
**2** - Views:
**1890**

### IMO 1974/1

Alice, Betty, and Carol took a series of examinations. There were one grade of $A$, one grade of $B$, and one grade of $C$ for each examination, where $A,B$ and $C$ are different positive integers. The final test scores were: Alice = $20$ Betty = $10$ Carol = $9$ If Betty placed first in the arithme...

- Mon Aug 24, 2015 10:24 pm
- Forum: National Math Camp
- Topic: Exam 1, Online Number Theory Camp, 2015
- Replies:
**15** - Views:
**9319**

### Re: Exam 1, Online Number Theory Camp, 2015

I am having the same problem......badass0 wrote:Please fix this

- Mon Jul 20, 2015 10:23 am
- Forum: Secondary Level
- Topic: Find XYZ
- Replies:
**10** - Views:
**4430**

### Re: Find XYZ

Got the proof:https://en.wikipedia.org/wiki/Markov_number.

So,there should be infinitively many solutions possible for this problem.

So,there should be infinitively many solutions possible for this problem.

- Mon Jul 20, 2015 1:08 am
- Forum: Secondary Level
- Topic: number theory
- Replies:
**1** - Views:
**1234**

### Re: number theory

We can observe that $$2014=2\times19\times53$$.Here $53$ is the biggest prime factor of $2014$. We have to find $\sum\frac{500}{53^n}$ for $n>0$. There are $9$ numbers that are multiples of $53$ from $1$ to $500$. No other multiples of $53^n$ are within numbers $1$ to $500$,where $n>1$. Thus,we get ...

- Fri Jul 10, 2015 7:48 pm
- Forum: Secondary Level
- Topic: 2014-national
- Replies:
**15** - Views:
**5912**

### Re: 2014-national

The solution is:

For an $m \times n$ chessboard, the number of the knights is $n \times \lceil\dfrac{m}{3}\rceil$.

For an $m \times n$ chessboard, the number of the knights is $n \times \lceil\dfrac{m}{3}\rceil$.

- Fri Jul 10, 2015 3:15 pm
- Forum: Secondary Level
- Topic: 2014-national
- Replies:
**15** - Views:
**5912**

### Re: 2014-national

You can place the knights in this order.This in an $8 \times 7$ chessboard.Hope this helps you.

- Tue Jul 07, 2015 9:09 pm
- Forum: Secondary Level
- Topic: Sylhet - 2014
- Replies:
**3** - Views:
**8134**

### Re: Sylhet - 2014

Since $x$ is divisible by $3k$, the maximum possible value of $3k$ will be $x$.

So, the maximum value if $k$ will be $x/3$.

We can also write it as $m^2(m+1)(m+2)(m+3)....(3m-1)$.

So, in this case the maximum value of $k$ will be $1000^2$x$1001$x$1002$......x$2999$.

So, the maximum value if $k$ will be $x/3$.

We can also write it as $m^2(m+1)(m+2)(m+3)....(3m-1)$.

So, in this case the maximum value of $k$ will be $1000^2$x$1001$x$1002$......x$2999$.

- Mon Jul 06, 2015 9:33 pm
- Forum: Secondary Level
- Topic: Regular Polygon inscribed in a circle
- Replies:
**5** - Views:
**3804**

### Re: Regular Polygon inscribed in a circle

A regular polygon inscribed in a circle with center $O$ has $n$ points(where $n$ is odd). Let, $$N=\lfloor\dfrac{n}{2}\rfloor$$. Then the number of the triangles having $O$ as inscribed is $N^2+(N-1)^2+(N-2)^2+........+1$. In this case the answer is $385$. Please let me know if there are any other t...

- Sun Jun 28, 2015 2:29 am
- Forum: Junior Level
- Topic: maximum area
- Replies:
**1** - Views:
**1600**

### Re: maximum area

When the perimeters are equal, the area of circle will be maximum. In this case while $N$ is the perimeter,for the triangles, the area of the equilateral triangle will be greater than the others.If $A$ is the area and $$s=(a+b+c)/2$$ of a triangle we can write it as $$A^2=s(s-a)(s-b)(s-c)$$.From the...