## Search found 57 matches

Thu Aug 29, 2019 9:07 pm
Forum: Algebra
Topic: 0^0=1!
Replies: 5
Views: 43112

### 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: 6523

### 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: 4731

### 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: 1351

### 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: 8163

### 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: 8163

### 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: 3959

### 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: 8163

### 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: 1375

### 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: 4749

### 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...