I found a sequence out of the serial multiplications, my result is $\frac{1001}{2000}$

I just observed the multiplications of first 7 numbers, please check it though

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- Thu Aug 25, 2011 2:23 am
- Forum: Secondary: Solved
- Topic: Dhaka Secondary 2009/4
- Replies:
**3** - Views:
**4532**

- Thu Aug 25, 2011 2:03 am
- Forum: Secondary: Solved
- Topic: Dhaka Secondary 2009/2
- Replies:
**6** - Views:
**5715**

### Re: Dhaka Secondary 2009/2

Moon vai I am posting a soultion,theoritical though,ar pura puri observation er upor prove kora,(observation gular nirghat prove ase but ami janina)so viewers please look for a better solution - ami amar koyekta observation mention kori( shobari egula jana so skip korleo chole):- 1.last digit 5 hole...

- Wed Aug 24, 2011 6:15 pm
- Forum: Computer Science
- Topic: solve it with turbo c
- Replies:
**21** - Views:
**7160**

### Re: solve it with turbo c

Ei program ta ki QBASIC diye kora jabe???

- Wed Aug 24, 2011 1:02 am
- Forum: H. Secondary: Solved
- Topic: Dhaka Higher Secondary 2010/10 (Secondary 2010/11)
- Replies:
**6** - Views:
**6321**

### Re: Dhaka Higher Secondary 2010/10 (Secondary 2010/11)

please give me a full solution, I am very weak in these sort of problems

- Tue Aug 23, 2011 1:19 pm
- Forum: Number Theory
- Topic: help please
- Replies:
**7** - Views:
**2388**

### Re: help please

I am posting a solution with induction:-Nine ten er higher math book follow kore dicchi. first as usual when n=1 then $5^{2n} + 3\times 2^{5n-2}$ is dividable by 7.......(1) now let for n=m $5^{2m} + 3\times 2^{5m-2}=7k$........(2) then induction law goes, if for n=m+1, $5^{2n} + 3\times 2^{5n-2}$ i...

- Tue Aug 23, 2011 12:38 pm
- Forum: Number Theory
- Topic: congruence problem again
- Replies:
**9** - Views:
**3024**

### Re: congruence problem again

solution 3- prove $x^{1005}\equiv 0, \pm1\pmod {2011}$ first if $x= 2011k$ then obviously $x^{1005} \equiv 0\pmod {2011}$ since $2011$ is prime and if $x\neq 2011k$ then $x$ and $2011$ are mutually co-prime so according to fermats little theorem $x^ {2011} \equiv x\pmod {2011}$ implies $ x^{2010}\eq...

- Tue Aug 23, 2011 3:11 am
- Forum: Number Theory
- Topic: congruence problem again
- Replies:
**9** - Views:
**3024**

### Re: congruence problem again

Since I am not used to modular arithmatics, I want to know a thing if $(a^n)^2 \equiv 1(mod p)$ then can we write $ (a^n)\equiv \pm1(mod p)$ ??? ( i have tried it on a number of numbers, they all seem to be true) if it can be written, then I have the solution for the first and the last one.....(plea...

- Mon Aug 22, 2011 9:25 pm
- Forum: Computer Science
- Topic: Is math essential for programming?
- Replies:
**11** - Views:
**6093**

### Re: Is math essential for programming?

I am sure u will understand this program to find the GCD. It is written in Qbasic so the easiest to get I suppose/ INPUT a, b DO x= a mod b let a=b let b=x LOOP UNTIL x=0 print a end Here "Loop until" means until the expression fill its demand(here as long as x=0) the process will be continued.... E...

- Mon Aug 22, 2011 8:49 pm
- Forum: Computer Science
- Topic: Programming Question
- Replies:
**6** - Views:
**7295**

### Re: Programming Question

Well I submit a way here, please check if it is programmable,

1.Input A

2.P=3

3. Compare- If (A-P) is a prime then N=A-P

print N,P. Goto 6

else go to next step

4. find the next prime form P to A

let the prime is Q

5. P=Q and goto 3

6.end

1.Input A

2.P=3

3. Compare- If (A-P) is a prime then N=A-P

print N,P. Goto 6

else go to next step

4. find the next prime form P to A

let the prime is Q

5. P=Q and goto 3

6.end

- Mon Aug 22, 2011 8:41 pm
- Forum: Computer Science
- Topic: Programming Question
- Replies:
**6** - Views:
**7295**

### Programming Question

I have a question to ask.... According to the conjecture of Goldbach we can write an even number (>2)..as the sum of two primes, example-8=5+3,90=83+7 etc...... Now I want to write a program using Qbasic (i am novice since).. where I will input an even number greater than 2, and the output will be t...