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: 9124
- Thu Aug 25, 2011 2:03 am
- Forum: Secondary: Solved
- Topic: Dhaka Secondary 2009/2
- Replies: 6
- Views: 10781
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: 12430
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: 11206
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: 4191
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: 5105
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: 5105
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: 10377
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: 13222
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: 13222
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...