## Search found 37 matches

- Mon Jan 07, 2019 9:01 pm
- Forum: Social Lounge
- Topic: How unfortunated we are!
- Replies:
**2** - Views:
**523**

### 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:
**379**

### 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:
**929**

### 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:
**691**

### 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:
**691**

### 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...

- Thu Sep 27, 2018 7:31 pm
- Forum: Algebra
- Topic: Minimum value
- Replies:
**3** - Views:
**523**

### Re: Minimum value

Apparently, the denominators should be 1. So the minimum value will be 6.

- Mon Sep 03, 2018 8:54 pm
- Forum: Higher Secondary Level
- Topic: Geometric question...Need help
- Replies:
**1** - Views:
**541**

### Re: Geometric question...Need help

I think the answer is 6.

- Mon Sep 03, 2018 1:42 am
- Forum: Secondary: Solved
- Topic: Divisional MO Secondary 2010
- Replies:
**2** - Views:
**1730**

### Re: Divisional MO Secondary 2010

Answer to ques. no. 12 is 201.

- Mon Sep 03, 2018 1:33 am
- Forum: Higher Secondary Level
- Topic: A problem of combinatorics
- Replies:
**9** - Views:
**1980**

### Re: A problem of combinatorics

Hint Use function Answer Correct answer $\fbox {3885}$ The correct answer of this problem is 546, and it is an established solution. This problem actually appeared in the National Math Olympiad 2015 in secondary and higher secondary categories. Moreover, read the question again. It says that no one...

- Mon Sep 03, 2018 1:31 am
- Forum: Higher Secondary Level
- Topic: A problem of combinatorics
- Replies:
**9** - Views:
**1980**

### Re: A problem of combinatorics

Read the problem again.