Anyone looking for a classic Graph Theory problem?

Look no further!!!

## Search found 62 matches

- Thu Oct 17, 2019 3:06 am
- Forum: International Mathematical Olympiad (IMO)
- Topic: IMO 2019/P3
- Replies:
**5** - Views:
**60256**

- Thu Oct 17, 2019 3:05 am
- Forum: International Mathematical Olympiad (IMO)
- Topic: IMO 2019/P3
- Replies:
**5** - Views:
**60256**

### Re: IMO 2019/P3

GRAPH THEORY... is printed all over this problem as a hint, isn't it?

- Thu Oct 03, 2019 1:43 am
- Forum: Combinatorics
- Topic: Binary Representation
- Replies:
**1** - Views:
**36571**

### Re: Binary Representation

Can you explain the RHS of the equation $B(nm) \geq \max{B(n),B(m)}$ ?

I mean, do we multiply or, add or, individually consider the maximum values of $B(n)$ and $B(m)$?

I mean, do we multiply or, add or, individually consider the maximum values of $B(n)$ and $B(m)$?

- Mon Sep 23, 2019 1:53 pm
- Forum: National Math Olympiad (BdMO)
- Topic: BdMO National Secondary 2019#6
- Replies:
**4** - Views:
**38628**

### Re: BdMO National Secondary 2019#6

Take one major diagonal and color the squares in black. Take the other two corners and color them in black. Take the diagonals consisting of 5 squares each (parallel to the colored major diagonal on either side) and color them in black. Therefore, total number of colored squares are $9+2+5+5=21$. It...

- Sun Sep 15, 2019 1:49 am
- Forum: Algebra
- Topic: Minimum value
- Replies:
**4** - Views:
**52980**

### Re: Minimum value

Apparently, the denominators should be 1. So the minimum value will be 6. I don't think so. Bcoz, x>0 and y>0 are real numbers not natural. I know the difference. What I meant was: in the solution to the problem, both $x$ and $y$ should be 1. BTW, this is just an observation since I didn't go throu...

- Mon Jan 07, 2019 9:01 pm
- Forum: Social Lounge
- Topic: How unfortunated we are!
- Replies:
**2** - Views:
**7763**

### Re: How unfortunated we are!

Of course we should. And you're right, this forum has turned into a 'death valley', which is very unfortunate. Once this forum was very active, with over 100 posts each day. But sometime in 2017 everybody got more involved with this forum called 'The Art of Problem Solving', and that was the beginni...

- Fri Oct 19, 2018 2:04 am
- Forum: National Math Olympiad (BdMO)
- Topic: BdMO National Secondary 2007/12
- Replies:
**2** - Views:
**1396**

### Re: BdMO National Secondary 2007/12

$x^2-1=0$

$x^2=1$

$x= +1$ or, $-1$

Let, $f(x)=x^{100}-2x^{51}+1$

Now, the remainder(s) can be found using the Remainder theorem.

When $x=+1$

$f(x)=1-2(1)+1$

$f(x)=0$

When $x=-1$

$f(x)=1-2(-1)+1$

$f(x)=4$

$x^2=1$

$x= +1$ or, $-1$

Let, $f(x)=x^{100}-2x^{51}+1$

Now, the remainder(s) can be found using the Remainder theorem.

When $x=+1$

$f(x)=1-2(1)+1$

$f(x)=0$

When $x=-1$

$f(x)=1-2(-1)+1$

$f(x)=4$

- Fri Oct 19, 2018 1:49 am
- Forum: Primary Level
- Topic: Combinatorics
- Replies:
**6** - Views:
**4867**

### Re: Combinatorics

The answer should be 14.

- Fri Oct 19, 2018 1:43 am
- Forum: Number Theory
- Topic: no solution (a,b)
- Replies:
**2** - Views:
**4398**

### Re: no solution (a,b)

There can be another solution to this problem, which includes a bit of messy work of Algebra. $a^2=b^7+7$ $a^2-16=b^7-9$ $(a+4)(a-4)=(\sqrt{b^7}+3)(\sqrt{b^7}-3)$ Now using a little bit of Algebra, it can be shown that there is no integer value of $a$ or, $b$ that satisfies the above equation, hence...

- Tue Oct 16, 2018 8:01 pm
- Forum: Number Theory
- Topic: no solution (a,b)
- Replies:
**2** - Views:
**4398**

### Re: no solution (a,b)

The LHS of the equation is always positive. So $b>0$, as for $b=0, a= \sqrt7$, and when $b=-1, a=\sqrt6$. And when $b<-1$, the equation becomes invalid. Therefore, it can be deduced that $b>0$. $a^2-b^7=7$ $(a+\sqrt{b^7}) (a-\sqrt{b^7})=7$ Now, $a>b$, otherwise the equation draws into a negative res...