## Search found 57 matches

### Re: 0^0=1!

You know the Binomial theorem: $ (x+y)^n = \sum_{k=0}^n {n \choose k} x^{n - k} y^k $ If one of the terms, say $y$, equals 0, then LHS $= (x + 0)^n = x^n$ and, on RHS, the 1st term of the expansion $={n \choose 0} x^n0^0 = x^n0^0$ and all the other terms become 0 because of $0^1, 0^2, 0^3, ... $. As...

- Mon Jun 24, 2019 7:51 pm
- Forum: Divisional Math Olympiad
- Topic: Mymensingh higher secondary 2013#1
- Replies:
**3** - Views:
**6426**

### Re: Mymensingh higher secondary 2013#1

There cannot be two

**at least**phrases in the problem to have a meaningful answer to it. One of these should be at least while the other**at best**.- Wed Jan 16, 2019 8:42 pm
- Forum: Primary Level
- Topic: Combinatorics
- Replies:
**6** - Views:
**4573**

### Re: Combinatorics

If repeated letters distract you, replace these in the following way, because the question basically asks you to choose and count properly: A B C D E F G H I J From A, you have 2 choices: AB and AC (one down and the other right-down diagonally). From each of AB and AC, you have also 2 choices: ABD, ...

- Fri Oct 26, 2018 3:47 pm
- Forum: National Math Olympiad (BdMO)
- Topic: BdMO National Secondary 2007/12
- Replies:
**2** - Views:
**1287**

### Re: BdMO National Secondary 2007/12

Since the highest power of the variable in your divisor polynomial $(x^2-1)$ is 2, your remainder polynomial will be in the form of $ax^1+b$. Right? Now suppose $f(x) = x^{100} - 2x^{51} + 1 = p(x)\cdot(x^2-1) + ax + b \qquad (Eq.1)$ You don't know about the quotient polynomial $p(x)$, and you don't...

- Mon Apr 23, 2018 5:36 pm
- Forum: National Math Olympiad (BdMO)
- Topic: BDMO 2017 National round Secondary 1
- Replies:
**19** - Views:
**7772**

### Re: BDMO 2017 National round Secondary 1

@ahmedittihad, I know where you have erred. In (c), you are calculating the probability of the series ending in 5 matches with the condition that Bangladesh must win the series. Nowhere in (c) is it mentioned that Bangladesh must win the series. About your invitation to a non-relevant forum, I have ...

- Tue Apr 10, 2018 9:14 pm
- Forum: National Math Olympiad (BdMO)
- Topic: BDMO 2017 National round Secondary 1
- Replies:
**19** - Views:
**7772**

### Re: BDMO 2017 National round Secondary 1

@ahmedittihad, I would rather say you are wrong. I have provided quite detailed clarification with easy-to-follow explanations. You have just drawn a conclusion without rebutting my arguments.

- Tue Apr 10, 2018 9:06 pm
- Forum: Physics
- Topic: will it come back
- Replies:
**9** - Views:
**3798**

### Re: will it come back

It’s not possible to define such a question without some inherent ambiguity in the question. The first ambiguity is “who is that ‘I’ who can throw a ball with mass amounting to the earth’s?” Even if you can clarify that person, he or she will have no time to throw, because the earth and the ball (pe...

- Thu Feb 08, 2018 1:35 am
- Forum: National Math Olympiad (BdMO)
- Topic: BDMO 2017 National round Secondary 1
- Replies:
**19** - Views:
**7772**

### Re: BDMO 2017 National round Secondary 1

The key to understanding this problem, as Ahmed pointed out, is that not all cases are equally likely to happen . The probabilities for the series to end in exactly 3 matches and to end in exactly 4 (or 5) matches are not equal, so the 20 cases listed above cannot be treated with equal weights. In g...

- Wed Nov 01, 2017 7:15 pm
- Forum: Secondary Level
- Topic: AREA
- Replies:
**1** - Views:
**1316**

### Re: AREA

Taking four pairs of sign for $(x, y)$ and considering magnitude of $x$ vs. that of $y$ in each of the four cases, you would get a square as the bounded region. It should then be a matter of moments to compute the area.

- Tue Aug 22, 2017 10:04 pm
- Forum: Algebra
- Topic: Double inequality
- Replies:
**3** - Views:
**4642**

### Re: Double inequality

There is a mistake in the variables of the problem. I've corrected $ x, y, z$ respectively to $ a, b, c $. We observe that $ bc + ca + ab - 2abc $ $ = bc - abc + ca - abc + ab -abc + abc $ $ = bc(1-a) + ca(1-b) + ab(1-c) + abc $ $ = bc(b+c) + ca(c+a) + ab(a+b) + abc \geq 0 $, because $ a, b, c $ eac...