## Search found 145 matches

- Sun Apr 11, 2021 6:31 pm
- Forum: Physics
- Topic: why so sun
- Replies:
**5** - Views:
**17**

### Re: why so sun

But not machines right? :?: Well, if the machine runs with solar power, then obviously yes. If it runs with oil, then that energy is coming from the energy stored by some life a few million years ago. If it uses energy that came from a nuclear power plant, then we can say it doesn't use sun power d...

- Sun Apr 11, 2021 12:53 pm
- Forum: Physics
- Topic: why so sun
- Replies:
**5** - Views:
**17**

### Re: why so sun

Is sun the "main" source of energy or "ultimate" source of energy? IF it is the main source of energy then why so? Cause the plants learnt how to convert the light energy to chemical energy. And each and almost every life on earth gets the energy from that chemical energy and convert them to some o...

- Sun Apr 11, 2021 12:27 pm
- Forum: Physics
- Topic: why so sun
- Replies:
**5** - Views:
**17**

### Re: why so sun

Is sun the "main" source of energy or "ultimate" source of energy? IF it is the main source of energy then why so? Cause the plants learnt how to convert the light energy to chemical energy. And each and almost every life on earth gets the energy from that chemical energy and convert them to some o...

- Sun Apr 11, 2021 11:11 am
- Forum: National Math Olympiad (BdMO)
- Topic: BdMO National 2021 Higher Secondary Problem 9
- Replies:
**2** - Views:
**144**

### Solution :

Let me start with an exercise. Find three distinct positive integers $a,b,c$ where $a<b<c$ such that \[\frac1a+\frac1b+\frac1c=1\] You may post the solution below. Well, the only solution to this equation is $(a,b,c)=(2,3,6)$. Let's assume, $n$ is a nice integer. Let $\frac{n}{x},\frac{n}{y},\frac{n...

- Sun Apr 11, 2021 10:31 am
- Forum: National Math Olympiad (BdMO)
- Topic: BdMO National 2021 Higher Secondary Problem 9
- Replies:
**2** - Views:
**144**

### BdMO National 2021 Higher Secondary Problem 9

A positive integer $n$ is called nice if it has at least $3$ proper divisors and it is equal to the sum of its three largest proper divisors. For example, $6$ is nice because its largest proper divisors are $3,2,1$ and $6=3+2+1$. Find the number of nice integers not greater than $6000$.

- Sat Apr 10, 2021 11:01 pm
- Forum: Algebra
- Topic: FE Marathon!
- Replies:
**96** - Views:
**4387**

### Re: Problem 25

Find functions $f:\mathbb{N}\to\mathbb{N}$ such that \[f(n)+f(f(n))+f(f(f(n)))=3n\ \ \ \forall n\in\mathbb{N}\] Where $\mathbb{N}$ is the set of all positive integers $\textbf{Solution 25}$ Its ez to see that its an injective function. Plugging $n=1 \ \Rightarrow \ f(1)+f(f(1))+f(f((1)))=3 \ $ $\Ri...

- Sat Apr 10, 2021 8:40 pm
- Forum: Social Lounge
- Topic: How was your national olympiad.
- Replies:
**6** - Views:
**37**

### Re: How was your national olympiad.

My exam:

আমি চিৎকার করে কাঁদিতে চাহিয়া করিতে পারি নি চিৎকার!

আমি চিৎকার করে কাঁদিতে চাহিয়া করিতে পারি নি চিৎকার!

- Sat Apr 10, 2021 8:25 pm
- Forum: National Math Olympiad (BdMO)
- Topic: BdMO National 2021 Higher Secondary Problem 7
- Replies:
**1** - Views:
**56**

### Re: Solution :

First thing you will notice is that $511=2^9-1$, which is the largest number with $9$ bit long binary string. For this particular problem, we will consider every integer number from $0$ to $2^9-1$ as a $9$-bit long binary string. For example, $2_{10}=10_2=00000010_2$, which is now a $9$ bit long bin...

- Sat Apr 10, 2021 4:54 pm
- Forum: National Math Olympiad (BdMO)
- Topic: BdMO National 2021 Higher Secondary Problem 7
- Replies:
**1** - Views:
**56**

### BdMO National 2021 Higher Secondary Problem 7

A binary string is a word containing only $0$s and $1$s. In a binary string, a $\text{1-run}$ is a non-extendable substring containing only $1$s. Given a positive number $n$, Let $B(n)$ be the number of $\text{1-runs}$ in the binary representation of $n$. For example, $B(107)=3$ since $107$ in binar...

- Thu Apr 08, 2021 12:51 pm
- Forum: Combinatorics
- Topic: A Combi problem
- Replies:
**3** - Views:
**58**

### Re: A Combi problem

If we transform the handshaking into a graph where the people are vertex and handshakes are represented by edges, we will get one or more disjoint cycles. Each cycle must have at least $3$ people so that everyone shakes hand with two other. So we have $4$ cases. We can take all of the people and pu...