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## Re: [math_club] alphabetical math (Ciphered Message)

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• Now I see!!! Interesting coincidence. Thank you very much for informing me this fact. I think from now on I should study about conjectures. Take care......
Message 1 of 16 , Oct 1 7:19 AM
Now I see!!! Interesting coincidence. Thank you very much for informing me this fact. I think from now on I should study about conjectures. Take care.....

From: Tanvir Prince <tanvirp123@...>
To: math_club@yahoogroups.com
Sent: Wednesday, September 30, 2009 11:10:13 PM
Subject: Re: [math_club] alphabetical math (Ciphered Message)

This conjecture is known as Levy's conjecture. Please read below:

In number theory, Lemoine's conjecture, named after Émile Lemoine, also known as Levy's conjecture, after Hyman Levy, states that all odd integers greater than 5 can be represented as the sum of an odd prime number and an even semiprime. To put it algebraically, 2n + 1 = p + 2q always has a solution in primes p and q (not necessarily distinct) for n > 2. The Lemoine conjecture is similar to but stronger than Goldbach's weak conjecture.
For example, 47 = 13 + 2 × 17 = 37 + 2 × 5 = 41 + 2 × 3 = 43 + 2 × 2. (sequence A046927 in OEIS) counts how many different ways 2n + 1 can be represented as p + 2q.
According to MathWorld, the conjecture has been verified by Corbitt up to 109.
The conjecture was posed by Émile Lemoine in 1895, but in more recent years came to be attributed to Hyman Levy who pondered it in the 1960s.

So as you can see the proof will be extreamly complicated since still it is a open question.

--- On Wed, 9/30/09, Mirza Sabbir Hossain Beg <mirzasabbirhossainb eg@yahoo. com> wrote:

From: Mirza Sabbir Hossain Beg <mirzasabbirhossainb eg@yahoo. com>
Subject: Re: [math_club] alphabetical math (Ciphered Message)
To: math_club@yahoogrou ps.com
Date: Wednesday, September 30, 2009, 3:40 AM

Dear Tanvir, if you think my assumption promising then share it and discuss with all the friends available in this club. Thanks for your mail........

From: Tanvir Prince <tanvirp123@yahoo. com>
To: math_club@yahoogrou ps.com
Sent: Tuesday, September 29, 2009 8:35:48 PM
Subject: Re: [math_club] alphabetical math (Ciphered Message)

Yes. you are absolutely right. I made a mistake. I was not that careful. This assumption looks very promisable to me although I do not have any proof.

--- On Tue, 9/29/09, Mirza Sabbir Hossain Beg <mirzasabbirhossainb eg@yahoo. com> wrote:

From: Mirza Sabbir Hossain Beg <mirzasabbirhossainb eg@yahoo. com>
Subject: Re: [math_club] alphabetical math (Ciphered Message)
To: math_club@yahoogrou ps..com
Date: Tuesday, September 29, 2009, 4:15 AM

Yes Tanvir, you made the mistake because 2*59 + 13 = 131, where both 59 and 13 are prime number. But pls check my assumption for a greater odd number, perhaps there could be a counter example. Above all I like to thank you for your attempt. Mail me if you can prove or disprove it.. Best regards..... ........

From: Tanvir Prince <tanvirp123@yahoo. com>
To: math_club@yahoogrou ps.com
Sent: Tuesday, September 29, 2009 2:31:36 AM
Subject: Re: [math_club] alphabetical math (Ciphered Message)

Your assumption that every odd number bigger than 5 can be written as 2p+q where p and q are prime numbers; I think I found a counter example which is 131.

can you write 131 = 2p + q where p and q prime.
It is possible that I make a mistake.
I use mathematica software to generate a list of numbers of the form 2p+q where p and q runs over all prime numbers starting from 2 to 71 and in that list it looks like 131 is missing..

--- On Mon, 9/28/09, Mirza Sabbir Hossain Beg <mirzasabbirhossainb eg@yahoo. com> wrote:

From: Mirza Sabbir Hossain Beg <mirzasabbirhossainb eg@yahoo. com>
Subject: Re: [math_club] alphabetical math (Ciphered Message)
To: math_club@yahoogrou ps.com
Date: Monday, September 28, 2009, 5:19 AM

Mr. Mamun, my assumption: 1. Any odd numbers greater than 5 can be written as this form: 2p + q, where p and q are prime numbers. 2.. Every prime numbers can be written as this form: 2(p+/-q) +/-1, where p and q are any prime numbers. Send me the proof or counter example if it has so. Thanks...... .......

From: S. M. Mamun Ar Rashid <mamun305@yahoo. com>
To: math_club@yahoogrou ps.com
Sent: Friday, September 25, 2009 10:10:55 PM
Subject: Re: [math_club] alphabetical math (Ciphered Message)

 The problem deals with a ciphered, i.e. encoded message. Your task is to decipher it, that is, to find out a pattern to see whether any meaningful English sentence reveals from it. One approach is to employ the Caesar Method [named after Julius Caesar, who used such technique of exchanging secret and sensitive messages] in which each letter is substituted by another letter. For example, aàc, bàd, càe. If it does not reveal anything, try aàd, bàe, càf. Hope you get the idea. The realm of Secret Message is very ancient, full of conundrums, mysterious, and often bloody.. One story might intrigue you. Mary, the queen of Scots, wanted to assassinate her cousin Queen Elizabeth I of England, and began exchanging secret, ciphered messages with her co-conspirators. But some of Mary's messages were captured by Elizabeth 's spies and after much effort they were cracked by her chief code breaker. Mary was immediately arrested on charge of treason and the deciphered messages were used in trial as evidence. She was found guilty and was beheaded in 1587. All because her cipher was cracked. Regards, Mamun--- On Fri, 9/25/09, mathandsnigdha wrote:From: mathandsnigdha Subject: [math_club] alphabetical mathTo: math_club@yahoogrou ps.comDate: Friday, September 25, 2009, 7:23 AM in "Neurone Anuranan" problem no-116 is disturbing me 4 a long time......i fail to catch it...... plz help me to find out its real meaning..... . ....

• Dear Tanvir, I also gave you another assumption that: any odd numbers can be expressed as this equation- 2(p+/-q)+/-1, where p and q are prime
Message 2 of 16 , Oct 1 8:11 AM
Dear Tanvir, I also gave you another assumption that: any odd numbers can be expressed as this equation- 2(p+/-q)+/-1, where p and q are prime numbers.(stronger than levy's and goldbach's conjecture!!!!!haha...ha just kidding man). Anyway, check and verify my assumption that it is unique or not. If it is unique then mail me proof or counter example. Mail me.

From: Tanvir Prince <tanvirp123@...>
To: math_club@yahoogroups.com
Sent: Wednesday, September 30, 2009 11:10:13 PM
Subject: Re: [math_club] alphabetical math (Ciphered Message)

This conjecture is known as Levy's conjecture. Please read below:

In number theory, Lemoine's conjecture, named after Émile Lemoine, also known as Levy's conjecture, after Hyman Levy, states that all odd integers greater than 5 can be represented as the sum of an odd prime number and an even semiprime. To put it algebraically, 2n + 1 = p + 2q always has a solution in primes p and q (not necessarily distinct) for n > 2. The Lemoine conjecture is similar to but stronger than Goldbach's weak conjecture.
For example, 47 = 13 + 2 × 17 = 37 + 2 × 5 = 41 + 2 × 3 = 43 + 2 × 2. (sequence A046927 in OEIS) counts how many different ways 2n + 1 can be represented as p + 2q.
According to MathWorld, the conjecture has been verified by Corbitt up to 109.
The conjecture was posed by Émile Lemoine in 1895, but in more recent years came to be attributed to Hyman Levy who pondered it in the 1960s.

So as you can see the proof will be extreamly complicated since still it is a open question.

--- On Wed, 9/30/09, Mirza Sabbir Hossain Beg <mirzasabbirhossainb eg@yahoo. com> wrote:

From: Mirza Sabbir Hossain Beg <mirzasabbirhossainb eg@yahoo. com>
Subject: Re: [math_club] alphabetical math (Ciphered Message)
To: math_club@yahoogrou ps.com
Date: Wednesday, September 30, 2009, 3:40 AM

Dear Tanvir, if you think my assumption promising then share it and discuss with all the friends available in this club. Thanks for your mail........

From: Tanvir Prince <tanvirp123@yahoo. com>
To: math_club@yahoogrou ps.com
Sent: Tuesday, September 29, 2009 8:35:48 PM
Subject: Re: [math_club] alphabetical math (Ciphered Message)

Yes. you are absolutely right. I made a mistake. I was not that careful. This assumption looks very promisable to me although I do not have any proof.

--- On Tue, 9/29/09, Mirza Sabbir Hossain Beg <mirzasabbirhossainb eg@yahoo. com> wrote:

From: Mirza Sabbir Hossain Beg <mirzasabbirhossainb eg@yahoo. com>
Subject: Re: [math_club] alphabetical math (Ciphered Message)
To: math_club@yahoogrou ps..com
Date: Tuesday, September 29, 2009, 4:15 AM

Yes Tanvir, you made the mistake because 2*59 + 13 = 131, where both 59 and 13 are prime number. But pls check my assumption for a greater odd number, perhaps there could be a counter example. Above all I like to thank you for your attempt. Mail me if you can prove or disprove it.. Best regards..... ........

From: Tanvir Prince <tanvirp123@yahoo. com>
To: math_club@yahoogrou ps.com
Sent: Tuesday, September 29, 2009 2:31:36 AM
Subject: Re: [math_club] alphabetical math (Ciphered Message)

Your assumption that every odd number bigger than 5 can be written as 2p+q where p and q are prime numbers; I think I found a counter example which is 131.

can you write 131 = 2p + q where p and q prime.
It is possible that I make a mistake.
I use mathematica software to generate a list of numbers of the form 2p+q where p and q runs over all prime numbers starting from 2 to 71 and in that list it looks like 131 is missing..

--- On Mon, 9/28/09, Mirza Sabbir Hossain Beg <mirzasabbirhossainb eg@yahoo. com> wrote:

From: Mirza Sabbir Hossain Beg <mirzasabbirhossainb eg@yahoo. com>
Subject: Re: [math_club] alphabetical math (Ciphered Message)
To: math_club@yahoogrou ps.com
Date: Monday, September 28, 2009, 5:19 AM

Mr. Mamun, my assumption: 1. Any odd numbers greater than 5 can be written as this form: 2p + q, where p and q are prime numbers. 2.. Every prime numbers can be written as this form: 2(p+/-q) +/-1, where p and q are any prime numbers. Send me the proof or counter example if it has so. Thanks...... .......

From: S. M. Mamun Ar Rashid <mamun305@yahoo. com>
To: math_club@yahoogrou ps.com
Sent: Friday, September 25, 2009 10:10:55 PM
Subject: Re: [math_club] alphabetical math (Ciphered Message)

 The problem deals with a ciphered, i.e. encoded message. Your task is to decipher it, that is, to find out a pattern to see whether any meaningful English sentence reveals from it. One approach is to employ the Caesar Method [named after Julius Caesar, who used such technique of exchanging secret and sensitive messages] in which each letter is substituted by another letter. For example, aàc, bàd, càe. If it does not reveal anything, try aàd, bàe, càf. Hope you get the idea. The realm of Secret Message is very ancient, full of conundrums, mysterious, and often bloody.. One story might intrigue you. Mary, the queen of Scots, wanted to assassinate her cousin Queen Elizabeth I of England, and began exchanging secret, ciphered messages with her co-conspirators. But some of Mary's messages were captured by Elizabeth 's spies and after much effort they were cracked by her chief code breaker. Mary was immediately arrested on charge of treason and the deciphered messages were used in trial as evidence. She was found guilty and was beheaded in 1587. All because her cipher was cracked. Regards, Mamun--- On Fri, 9/25/09, mathandsnigdha wrote:From: mathandsnigdha Subject: [math_club] alphabetical mathTo: math_club@yahoogrou ps.comDate: Friday, September 25, 2009, 7:23 AM in "Neurone Anuranan" problem no-116 is disturbing me 4 a long time......i fail to catch it...... plz help me to find out its real meaning..... . ....

• This assumption can be proven using Goldbach s conjecture but again technically it is not a prove since it depends on Goldbach s conjecture which has not been
Message 3 of 16 , Oct 1 1:43 PM
This assumption can be proven using Goldbach's conjecture but again technically it is not a prove since it depends on Goldbach's conjecture which has not been proven yet.

First note that any odd number can be written as 2(n) +/- 1 where n is even. The reason is as follows:
First any odd number can be written 2(n) - 1; if n is already even then we are done. if n is odd then simply write 2(n-1) + 1 because 2(n) + 1 = 2(n-1) + 1 and now n-1 must be even.

On the other hand Goldbach's conjecture said that any even number can be written as a sum of two prime; so we can write our n as n = p+q where p and q are prime. This shows any odd number can be written as 2(p+q) +/- 1. And this finishes the proof of your conjecture. This is infact stronger than your original assumption since your original assumption is 2(p+/- q) +/- 1.

--- On Thu, 10/1/09, Mirza Sabbir Hossain Beg <mirzasabbirhossainbeg@...> wrote:

From: Mirza Sabbir Hossain Beg <mirzasabbirhossainbeg@...>
Subject: Re: [math_club] alphabetical math (Ciphered Message)
To: math_club@yahoogroups.com
Date: Thursday, October 1, 2009, 11:11 AM

Dear Tanvir, I also gave you another assumption that: any odd numbers can be expressed as this equation- 2(p+/-q)+/-1, where p and q are prime numbers.(stronger than levy's and goldbach's conjecture!! !!!haha.. .ha just kidding man). Anyway, check and verify my assumption that it is unique or not. If it is unique then mail me proof or counter example. Mail me.

From: Tanvir Prince <tanvirp123@yahoo. com>
To: math_club@yahoogrou ps.com
Sent: Wednesday, September 30, 2009 11:10:13 PM
Subject: Re: [math_club] alphabetical math (Ciphered Message)

This conjecture is known as Levy's conjecture. Please read below:

In number theory, Lemoine's conjecture, named after Émile Lemoine, also known as Levy's conjecture, after Hyman Levy, states that all odd integers greater than 5 can be represented as the sum of an odd prime number and an even semiprime. To put it algebraically, 2n + 1 = p + 2q always has a solution in primes p and q (not necessarily distinct) for n > 2. The Lemoine conjecture is similar to but stronger than Goldbach's weak conjecture.
For example, 47 = 13 + 2 × 17 = 37 + 2 × 5 = 41 + 2 × 3 = 43 + 2 × 2. (sequence A046927 in OEIS) counts how many different ways 2n + 1 can be represented as p + 2q.
According to MathWorld, the conjecture has been verified by Corbitt up to 109.
The conjecture was posed by Émile Lemoine in 1895, but in more recent years came to be attributed to Hyman Levy who pondered it in the 1960s.

So as you can see the proof will be extreamly complicated since still it is a open question.

--- On Wed, 9/30/09, Mirza Sabbir Hossain Beg <mirzasabbirhossainb eg@yahoo. com> wrote:

From: Mirza Sabbir Hossain Beg <mirzasabbirhossainb eg@yahoo. com>
Subject: Re: [math_club] alphabetical math (Ciphered Message)
To: math_club@yahoogrou ps.com
Date: Wednesday, September 30, 2009, 3:40 AM

Dear Tanvir, if you think my assumption promising then share it and discuss with all the friends available in this club. Thanks for your mail........

From: Tanvir Prince <tanvirp123@yahoo. com>
To: math_club@yahoogrou ps.com
Sent: Tuesday, September 29, 2009 8:35:48 PM
Subject: Re: [math_club] alphabetical math (Ciphered Message)

Yes. you are absolutely right. I made a mistake. I was not that careful. This assumption looks very promisable to me although I do not have any proof.

--- On Tue, 9/29/09, Mirza Sabbir Hossain Beg <mirzasabbirhossainb eg@yahoo. com> wrote:

From: Mirza Sabbir Hossain Beg <mirzasabbirhossainb eg@yahoo. com>
Subject: Re: [math_club] alphabetical math (Ciphered Message)
To: math_club@yahoogrou ps..com
Date: Tuesday, September 29, 2009, 4:15 AM

Yes Tanvir, you made the mistake because 2*59 + 13 = 131, where both 59 and 13 are prime number. But pls check my assumption for a greater odd number, perhaps there could be a counter example. Above all I like to thank you for your attempt. Mail me if you can prove or disprove it.. Best regards..... ........

From: Tanvir Prince <tanvirp123@yahoo. com>
To: math_club@yahoogrou ps.com
Sent: Tuesday, September 29, 2009 2:31:36 AM
Subject: Re: [math_club] alphabetical math (Ciphered Message)

Your assumption that every odd number bigger than 5 can be written as 2p+q where p and q are prime numbers; I think I found a counter example which is 131.

can you write 131 = 2p + q where p and q prime.
It is possible that I make a mistake.
I use mathematica software to generate a list of numbers of the form 2p+q where p and q runs over all prime numbers starting from 2 to 71 and in that list it looks like 131 is missing..

--- On Mon, 9/28/09, Mirza Sabbir Hossain Beg <mirzasabbirhossainb eg@yahoo. com> wrote:

From: Mirza Sabbir Hossain Beg <mirzasabbirhossainb eg@yahoo. com>
Subject: Re: [math_club] alphabetical math (Ciphered Message)
To: math_club@yahoogrou ps.com
Date: Monday, September 28, 2009, 5:19 AM

Mr. Mamun, my assumption: 1. Any odd numbers greater than 5 can be written as this form: 2p + q, where p and q are prime numbers. 2.. Every prime numbers can be written as this form: 2(p+/-q) +/-1, where p and q are any prime numbers. Send me the proof or counter example if it has so. Thanks...... .......

From: S. M. Mamun Ar Rashid <mamun305@yahoo. com>
To: math_club@yahoogrou ps.com
Sent: Friday, September 25, 2009 10:10:55 PM
Subject: Re: [math_club] alphabetical math (Ciphered Message)

 The problem deals with a ciphered, i.e. encoded message. Your task is to decipher it, that is, to find out a pattern to see whether any meaningful English sentence reveals from it. One approach is to employ the Caesar Method [named after Julius Caesar, who used such technique of exchanging secret and sensitive messages] in which each letter is substituted by another letter. For example, aàc, bàd, càe. If it does not reveal anything, try aàd, bàe, càf. Hope you get the idea. The realm of Secret Message is very ancient, full of conundrums, mysterious, and often bloody.. One story might intrigue you. Mary, the queen of Scots, wanted to assassinate her cousin Queen Elizabeth I of England, and began exchanging secret, ciphered messages with her co-conspirators. But some of Mary's messages were captured by Elizabeth 's spies and after much effort they were cracked by her chief code breaker. Mary was immediately arrested on charge of treason and the deciphered messages were used in trial as evidence. She was found guilty and was beheaded in 1587. All because her cipher was cracked. Regards, Mamun--- On Fri, 9/25/09, mathandsnigdha wrote:From: mathandsnigdha Subject: [math_club] alphabetical mathTo: math_club@yahoogrou ps.comDate: Friday, September 25, 2009, 7:23 AM in "Neurone Anuranan" problem no-116 is disturbing me 4 a long time......i fail to catch it...... plz help me to find out its real meaning..... . ....

• Thanks for the mail but I think you did some mistakes! First you said that any odd number can be written as 2(n) + / - 1 where n is even but the fact is n must
Message 4 of 16 , Oct 2 1:37 AM
Thanks for the mail but I think you did some mistakes! First you said that any odd number can be written as 2(n) + / - 1 where n is even but the fact is n must be odd also, otherwise this equation gives odd numbers in this progression: (+ / -) = 3, 5, 7, 9, 11,13.......17,19....... where 1 is always missing so the correct equation for generating odd numbers is 2n - 1 = 1, 3, 5, 7.......11, 13, 23, 25............37, 39..where n is the set of all natural numbers with the perfect sequence: 1, 2, 3, 4, 5, 6, 7, 8, 9........ so the equation 2n - 1 produces odd numbers infinitely. In the time of Goldbach the number 1 is considered as a prime number but then in the age of modern mathematics 1 is considered as the unit number!!!!! Why? Because if not then the theory of prime factorizations loses its uniqueness!! You are encouraged to study about this. So the ultimate modification and correction of Goldbach conjecture is: "any even number greater 4 can be written as the sum of two prime numbers and any odd numbers greater than 9 can be written as the sum of three prime numbers." So my conjecture is not proved. Because 2(3 - 2) - 1 = 1 is the only solution for p and q are prime. So, 2(p+q) + / -1 is not equal to 1 even p and q are prime. Your proposed equation is unable to generate 3 and 5 also because the lowest interval for p and q are prime: 2(2 + 2) = 8 + / - 1 = (7,9). Where is 3 and 5??!!!!?!! So your equation and proof is false. My equation still holds its correctness. So as far as the levy's conjecture went up to 5 my equation holds its correctness for 1, 3, 5 and rest of the odd numbers. So if you don't have any counter example of my assumption yet, than my assumption is still stronger than the Levy's and Goldbach's. Mail me............

From: Tanvir Prince <tanvirp123@...>
To: math_club@yahoogroups.com
Cc: tanvirp123@...
Sent: Friday, October 2, 2009 3:43:38 AM
Subject: Re: [math_club] alphabetical math (Ciphered Message)

This assumption can be proven using Goldbach's conjecture but again technically it is not a prove since it depends on Goldbach's conjecture which has not been proven yet.

First note that any odd number can be written as 2(n) +/- 1 where n is even. The reason is as follows:
First any odd number can be written 2(n) - 1; if n is already even then we are done. if n is odd then simply write 2(n-1) + 1 because 2(n) + 1 = 2(n-1) + 1 and now n-1 must be even.

On the other hand Goldbach's conjecture said that any even number can be written as a sum of two prime; so we can write our n as n = p+q where p and q are prime. This shows any odd number can be written as 2(p+q) +/- 1. And this finishes the proof of your conjecture. This is infact stronger than your original assumption since your original assumption is 2(p+/- q) +/- 1.

--- On Thu, 10/1/09, Mirza Sabbir Hossain Beg <mirzasabbirhossainb eg@yahoo. com> wrote:

From: Mirza Sabbir Hossain Beg <mirzasabbirhossainb eg@yahoo. com>
Subject: Re: [math_club] alphabetical math (Ciphered Message)
To: math_club@yahoogrou ps.com
Date: Thursday, October 1, 2009, 11:11 AM

Dear Tanvir, I also gave you another assumption that: any odd numbers can be expressed as this equation- 2(p+/-q)+/-1, where p and q are prime numbers.(stronger than levy's and goldbach's conjecture!! !!!haha.. .ha just kidding man). Anyway, check and verify my assumption that it is unique or not. If it is unique then mail me proof or counter example. Mail me.

From: Tanvir Prince <tanvirp123@yahoo. com>
To: math_club@yahoogrou ps.com
Sent: Wednesday, September 30, 2009 11:10:13 PM
Subject: Re: [math_club] alphabetical math (Ciphered Message)

This conjecture is known as Levy's conjecture. Please read below:

In number theory, Lemoine's conjecture, named after Émile Lemoine, also known as Levy's conjecture, after Hyman Levy, states that all odd integers greater than 5 can be represented as the sum of an odd prime number and an even semiprime. To put it algebraically, 2n + 1 = p + 2q always has a solution in primes p and q (not necessarily distinct) for n > 2. The Lemoine conjecture is similar to but stronger than Goldbach's weak conjecture.
For example, 47 = 13 + 2 × 17 = 37 + 2 × 5 = 41 + 2 × 3 = 43 + 2 × 2. (sequence A046927 in OEIS) counts how many different ways 2n + 1 can be represented as p + 2q.
According to MathWorld, the conjecture has been verified by Corbitt up to 109.
The conjecture was posed by Émile Lemoine in 1895, but in more recent years came to be attributed to Hyman Levy who pondered it in the 1960s.

So as you can see the proof will be extreamly complicated since still it is a open question.

--- On Wed, 9/30/09, Mirza Sabbir Hossain Beg <mirzasabbirhossainb eg@yahoo. com> wrote:

From: Mirza Sabbir Hossain Beg <mirzasabbirhossainb eg@yahoo. com>
Subject: Re: [math_club] alphabetical math (Ciphered Message)
To: math_club@yahoogrou ps.com
Date: Wednesday, September 30, 2009, 3:40 AM

Dear Tanvir, if you think my assumption promising then share it and discuss with all the friends available in this club. Thanks for your mail........

From: Tanvir Prince <tanvirp123@yahoo. com>
To: math_club@yahoogrou ps.com
Sent: Tuesday, September 29, 2009 8:35:48 PM
Subject: Re: [math_club] alphabetical math (Ciphered Message)

Yes. you are absolutely right. I made a mistake. I was not that careful. This assumption looks very promisable to me although I do not have any proof.

--- On Tue, 9/29/09, Mirza Sabbir Hossain Beg <mirzasabbirhossainb eg@yahoo. com> wrote:

From: Mirza Sabbir Hossain Beg <mirzasabbirhossainb eg@yahoo. com>
Subject: Re: [math_club] alphabetical math (Ciphered Message)
To: math_club@yahoogrou ps...com
Date: Tuesday, September 29, 2009, 4:15 AM

Yes Tanvir, you made the mistake because 2*59 + 13 = 131, where both 59 and 13 are prime number. But pls check my assumption for a greater odd number, perhaps there could be a counter example. Above all I like to thank you for your attempt. Mail me if you can prove or disprove it.. Best regards..... ........

From: Tanvir Prince <tanvirp123@yahoo. com>
To: math_club@yahoogrou ps.com
Sent: Tuesday, September 29, 2009 2:31:36 AM
Subject: Re: [math_club] alphabetical math (Ciphered Message)

Your assumption that every odd number bigger than 5 can be written as 2p+q where p and q are prime numbers; I think I found a counter example which is 131.

can you write 131 = 2p + q where p and q prime.
It is possible that I make a mistake.
I use mathematica software to generate a list of numbers of the form 2p+q where p and q runs over all prime numbers starting from 2 to 71 and in that list it looks like 131 is missing..

--- On Mon, 9/28/09, Mirza Sabbir Hossain Beg <mirzasabbirhossainb eg@yahoo. com> wrote:

From: Mirza Sabbir Hossain Beg <mirzasabbirhossainb eg@yahoo. com>
Subject: Re: [math_club] alphabetical math (Ciphered Message)
To: math_club@yahoogrou ps.com
Date: Monday, September 28, 2009, 5:19 AM

Mr. Mamun, my assumption: 1. Any odd numbers greater than 5 can be written as this form: 2p + q, where p and q are prime numbers. 2... Every prime numbers can be written as this form: 2(p+/-q) +/-1, where p and q are any prime numbers. Send me the proof or counter example if it has so. Thanks...... .......

From: S. M. Mamun Ar Rashid <mamun305@yahoo. com>
To: math_club@yahoogrou ps.com
Sent: Friday, September 25, 2009 10:10:55 PM
Subject: Re: [math_club] alphabetical math (Ciphered Message)

 The problem deals with a ciphered, i.e. encoded message. Your task is to decipher it, that is, to find out a pattern to see whether any meaningful English sentence reveals from it. One approach is to employ the Caesar Method [named after Julius Caesar, who used such technique of exchanging secret and sensitive messages] in which each letter is substituted by another letter. For example, aàc, bàd, càe. If it does not reveal anything, try aàd, bàe, càf. Hope you get the idea. The realm of Secret Message is very ancient, full of conundrums, mysterious, and often bloody.. One story might intrigue you. Mary, the queen of Scots, wanted to assassinate her cousin Queen Elizabeth I of England, and began exchanging secret, ciphered messages with her co-conspirators. But some of Mary's messages were captured by Elizabeth 's spies and after much effort they were cracked by her chief code breaker. Mary was immediately arrested on charge of treason and the deciphered messages were used in trial as evidence. She was found guilty and was beheaded in 1587. All because her cipher was cracked. Regards, Mamun--- On Fri, 9/25/09, mathandsnigdha wrote:From: mathandsnigdha Subject: [math_club] alphabetical mathTo: math_club@yahoogrou ps.comDate: Friday, September 25, 2009, 7:23 AM in "Neurone Anuranan" problem no-116 is disturbing me 4 a long time......i fail to catch it...... plz help me to find out its real meaning..... . .....

•   To be more precise, may be I should write the proof this way: first any odd number = 2 (n) +/- 1 ,where n is even, is still true since 0 is also even and 1
Message 5 of 16 , Oct 2 5:32 AM

To be more precise, may be I should write the proof this way:
first any odd number = 2 (n) +/- 1 ,where n is even, is still true since 0 is also even and 1 = 2 (0) +1 but this is not actually the important point I want to make.
Now except for the first few odd numbers which can be checked by hand:
1 = 2(3 - 3 ) +1
3 = 2(3 - 2) +1
5 = 2(5-3 ) +1

the rest of the odd number can written 2 ( p+ q) +/- 1

In any case there is no doubt that the proof before at lease prove your conjecture right?
since it show for the first three odd number we can write 2( p - q) +1 and for the rest
2(p+q) +/-1. so together we have any odd number can be written 2( p +/- q ) +/- 1 which is your original conjecture.

--- On Fri, 10/2/09, Mirza Sabbir Hossain Beg <mirzasabbirhossainbeg@...> wrote:

From: Mirza Sabbir Hossain Beg <mirzasabbirhossainbeg@...>
Subject: Re: [math_club] alphabetical math (Ciphered Message)
To: math_club@yahoogroups.com
Date: Friday, October 2, 2009, 4:37 AM

Thanks for the mail but I think you did some mistakes! First you said that any odd number can be written as 2(n) + / - 1 where n is even but the fact is n must be odd also, otherwise this equation gives odd numbers in this progression: (+ / -) = 3, 5, 7, 9, 11,13....... 17,19.... ... where 1 is always missing so the correct equation for generating odd numbers is 2n - 1 = 1, 3, 5, 7.......11, 13, 23, 25.......... ..37, 39..where n is the set of all natural numbers with the perfect sequence: 1, 2, 3, 4, 5, 6, 7, 8, 9........ so the equation 2n - 1 produces odd numbers infinitely. In the time of Goldbach the number 1 is considered as a prime number but then in the age of modern mathematics 1 is considered as the unit number!!!!! Why? Because if not then the theory of prime factorizations loses its uniqueness!! You are encouraged to study about this. So the ultimate modification and correction of Goldbach conjecture is: "any even number greater 4 can be written as the sum of two prime numbers and any odd numbers greater than 9 can be written as the sum of three prime numbers." So my conjecture is not proved. Because 2(3 - 2) - 1 = 1 is the only solution for p and q are prime. So, 2(p+q) + / -1 is not equal to 1 even p and q are prime. Your proposed equation is unable to generate 3 and 5 also because the lowest interval for p and q are prime: 2(2 + 2) = 8 + / - 1 = (7,9). Where is 3 and 5??!!!!?!! So your equation and proof is false. My equation still holds its correctness. So as far as the levy's conjecture went up to 5 my equation holds its correctness for 1, 3, 5 and rest of the odd numbers. So if you don't have any counter example of my assumption yet, than my assumption is still stronger than the Levy's and Goldbach's. Mail me.......... ..

From: Tanvir Prince <tanvirp123@yahoo. com>
To: math_club@yahoogrou ps.com
Cc: tanvirp123@yahoo. com
Sent: Friday, October 2, 2009 3:43:38 AM
Subject: Re: [math_club] alphabetical math (Ciphered Message)

This assumption can be proven using Goldbach's conjecture but again technically it is not a prove since it depends on Goldbach's conjecture which has not been proven yet.

First note that any odd number can be written as 2(n) +/- 1 where n is even. The reason is as follows:
First any odd number can be written 2(n) - 1; if n is already even then we are done. if n is odd then simply write 2(n-1) + 1 because 2(n) + 1 = 2(n-1) + 1 and now n-1 must be even.

On the other hand Goldbach's conjecture said that any even number can be written as a sum of two prime; so we can write our n as n = p+q where p and q are prime. This shows any odd number can be written as 2(p+q) +/- 1. And this finishes the proof of your conjecture. This is infact stronger than your original assumption since your original assumption is 2(p+/- q) +/- 1.

--- On Thu, 10/1/09, Mirza Sabbir Hossain Beg <mirzasabbirhossainb eg@yahoo. com> wrote:

From: Mirza Sabbir Hossain Beg <mirzasabbirhossainb eg@yahoo. com>
Subject: Re: [math_club] alphabetical math (Ciphered Message)
To: math_club@yahoogrou ps.com
Date: Thursday, October 1, 2009, 11:11 AM

Dear Tanvir, I also gave you another assumption that: any odd numbers can be expressed as this equation- 2(p+/-q)+/-1, where p and q are prime numbers.(stronger than levy's and goldbach's conjecture!! !!!haha.. .ha just kidding man). Anyway, check and verify my assumption that it is unique or not. If it is unique then mail me proof or counter example. Mail me.

From: Tanvir Prince <tanvirp123@yahoo. com>
To: math_club@yahoogrou ps.com
Sent: Wednesday, September 30, 2009 11:10:13 PM
Subject: Re: [math_club] alphabetical math (Ciphered Message)

This conjecture is known as Levy's conjecture. Please read below:

In number theory, Lemoine's conjecture, named after Émile Lemoine, also known as Levy's conjecture, after Hyman Levy, states that all odd integers greater than 5 can be represented as the sum of an odd prime number and an even semiprime. To put it algebraically, 2n + 1 = p + 2q always has a solution in primes p and q (not necessarily distinct) for n > 2. The Lemoine conjecture is similar to but stronger than Goldbach's weak conjecture.
For example, 47 = 13 + 2 × 17 = 37 + 2 × 5 = 41 + 2 × 3 = 43 + 2 × 2. (sequence A046927 in OEIS) counts how many different ways 2n + 1 can be represented as p + 2q.
According to MathWorld, the conjecture has been verified by Corbitt up to 109.
The conjecture was posed by Émile Lemoine in 1895, but in more recent years came to be attributed to Hyman Levy who pondered it in the 1960s.

So as you can see the proof will be extreamly complicated since still it is a open question.

--- On Wed, 9/30/09, Mirza Sabbir Hossain Beg <mirzasabbirhossainb eg@yahoo. com> wrote:

From: Mirza Sabbir Hossain Beg <mirzasabbirhossainb eg@yahoo. com>
Subject: Re: [math_club] alphabetical math (Ciphered Message)
To: math_club@yahoogrou ps.com
Date: Wednesday, September 30, 2009, 3:40 AM

Dear Tanvir, if you think my assumption promising then share it and discuss with all the friends available in this club. Thanks for your mail........

From: Tanvir Prince <tanvirp123@yahoo. com>
To: math_club@yahoogrou ps.com
Sent: Tuesday, September 29, 2009 8:35:48 PM
Subject: Re: [math_club] alphabetical math (Ciphered Message)

Yes. you are absolutely right. I made a mistake. I was not that careful. This assumption looks very promisable to me although I do not have any proof.

--- On Tue, 9/29/09, Mirza Sabbir Hossain Beg <mirzasabbirhossainb eg@yahoo. com> wrote:

From: Mirza Sabbir Hossain Beg <mirzasabbirhossainb eg@yahoo. com>
Subject: Re: [math_club] alphabetical math (Ciphered Message)
To: math_club@yahoogrou ps...com
Date: Tuesday, September 29, 2009, 4:15 AM

Yes Tanvir, you made the mistake because 2*59 + 13 = 131, where both 59 and 13 are prime number. But pls check my assumption for a greater odd number, perhaps there could be a counter example. Above all I like to thank you for your attempt. Mail me if you can prove or disprove it.. Best regards..... ........

From: Tanvir Prince <tanvirp123@yahoo. com>
To: math_club@yahoogrou ps.com
Sent: Tuesday, September 29, 2009 2:31:36 AM
Subject: Re: [math_club] alphabetical math (Ciphered Message)

Your assumption that every odd number bigger than 5 can be written as 2p+q where p and q are prime numbers; I think I found a counter example which is 131.

can you write 131 = 2p + q where p and q prime.
It is possible that I make a mistake.
I use mathematica software to generate a list of numbers of the form 2p+q where p and q runs over all prime numbers starting from 2 to 71 and in that list it looks like 131 is missing..

--- On Mon, 9/28/09, Mirza Sabbir Hossain Beg <mirzasabbirhossainb eg@yahoo. com> wrote:

From: Mirza Sabbir Hossain Beg <mirzasabbirhossainb eg@yahoo. com>
Subject: Re: [math_club] alphabetical math (Ciphered Message)
To: math_club@yahoogrou ps.com
Date: Monday, September 28, 2009, 5:19 AM

Mr. Mamun, my assumption: 1. Any odd numbers greater than 5 can be written as this form: 2p + q, where p and q are prime numbers. 2... Every prime numbers can be written as this form: 2(p+/-q) +/-1, where p and q are any prime numbers. Send me the proof or counter example if it has so. Thanks...... .......

From: S. M. Mamun Ar Rashid <mamun305@yahoo. com>
To: math_club@yahoogrou ps.com
Sent: Friday, September 25, 2009 10:10:55 PM
Subject: Re: [math_club] alphabetical math (Ciphered Message)

 The problem deals with a ciphered, i.e. encoded message. Your task is to decipher it, that is, to find out a pattern to see whether any meaningful English sentence reveals from it. One approach is to employ the Caesar Method [named after Julius Caesar, who used such technique of exchanging secret and sensitive messages] in which each letter is substituted by another letter. For example, aàc, bàd, càe. If it does not reveal anything, try aàd, bàe, càf. Hope you get the idea. The realm of Secret Message is very ancient, full of conundrums, mysterious, and often bloody.. One story might intrigue you. Mary, the queen of Scots, wanted to assassinate her cousin Queen Elizabeth I of England, and began exchanging secret, ciphered messages with her co-conspirators. But some of Mary's messages were captured by Elizabeth 's spies and after much effort they were cracked by her chief code breaker. Mary was immediately arrested on charge of treason and the deciphered messages were used in trial as evidence. She was found guilty and was beheaded in 1587. All because her cipher was cracked. Regards, Mamun--- On Fri, 9/25/09, mathandsnigdha wrote:From: mathandsnigdha Subject: [math_club] alphabetical mathTo: math_club@yahoogrou ps.comDate: Friday, September 25, 2009, 7:23 AM in "Neurone Anuranan" problem no-116 is disturbing me 4 a long time......i fail to catch it...... plz help me to find out its real meaning..... . .....

• Yes Tanvir you got the point but all you did is give some examples that obey the rule of my assumption. But the fact is a counter example is enough to disprove
Message 6 of 16 , Oct 2 8:48 AM
Yes Tanvir you got the point but all you did is give some examples that obey the rule of my assumption. But the fact is a counter example is enough to disprove an assumption but an example cannot be the proof of an assumption. You have to give "elemental proof". You showed that 1, 3, 5 can be written as 2(p+/-q)+/-1, but the others? Not necessarily, because they can be written in a more consistent way [2(p+q)+/-1]. I agree. So that is the strongest point of my assumption because it coveres all the odd integers[2(p+/-q)+/-1]. I think now you understand my point. However, here is my complete assumption: every even integer can be written as 2(p+/-q)+/-2 and every odd integers can be written as 2(p+/-q)+/-1, where p and q are prime numbers.

From: Tanvir Prince <tanvirp123@...>
To: math_club@yahoogroups.com
Sent: Friday, October 2, 2009 7:32:51 PM
Subject: Re: [math_club] alphabetical math (Ciphered Message)

To be more precise, may be I should write the proof this way:
first any odd number = 2 (n) +/- 1 ,where n is even, is still true since 0 is also even and 1 = 2 (0) +1 but this is not actually the important point I want to make.
Now except for the first few odd numbers which can be checked by hand:
1 = 2(3 - 3 ) +1
3 = 2(3 - 2) +1
5 = 2(5-3 ) +1

the rest of the odd number can written 2 ( p+ q) +/- 1

In any case there is no doubt that the proof before at lease prove your conjecture right?
since it show for the first three odd number we can write 2( p - q) +1 and for the rest
2(p+q) +/-1. so together we have any odd number can be written 2( p +/- q ) +/- 1 which is your original conjecture.

--- On Fri, 10/2/09, Mirza Sabbir Hossain Beg <mirzasabbirhossainb eg@yahoo. com> wrote:

From: Mirza Sabbir Hossain Beg <mirzasabbirhossainb eg@yahoo. com>
Subject: Re: [math_club] alphabetical math (Ciphered Message)
To: math_club@yahoogrou ps.com
Date: Friday, October 2, 2009, 4:37 AM

Thanks for the mail but I think you did some mistakes! First you said that any odd number can be written as 2(n) + / - 1 where n is even but the fact is n must be odd also, otherwise this equation gives odd numbers in this progression: (+ / -) = 3, 5, 7, 9, 11,13....... 17,19..... ... where 1 is always missing so the correct equation for generating odd numbers is 2n - 1 = 1, 3, 5, 7.......11, 13, 23, 25.......... ..37, 39..where n is the set of all natural numbers with the perfect sequence: 1, 2, 3, 4, 5, 6, 7, 8, 9........ so the equation 2n - 1 produces odd numbers infinitely. In the time of Goldbach the number 1 is considered as a prime number but then in the age of modern mathematics 1 is considered as the unit number!!!!! Why? Because if not then the theory of prime factorizations loses its uniqueness!! You are encouraged to study about this. So the ultimate modification and correction of Goldbach conjecture is: "any even number greater 4 can be written as the sum of two prime numbers and any odd numbers greater than 9 can be written as the sum of three prime numbers." So my conjecture is not proved. Because 2(3 - 2) - 1 = 1 is the only solution for p and q are prime. So, 2(p+q) + / -1 is not equal to 1 even p and q are prime. Your proposed equation is unable to generate 3 and 5 also because the lowest interval for p and q are prime: 2(2 + 2) = 8 + / - 1 = (7,9). Where is 3 and 5??!!!!?!! So your equation and proof is false. My equation still holds its correctness. So as far as the levy's conjecture went up to 5 my equation holds its correctness for 1, 3, 5 and rest of the odd numbers. So if you don't have any counter example of my assumption yet, than my assumption is still stronger than the Levy's and Goldbach's. Mail me.......... ...

From: Tanvir Prince <tanvirp123@yahoo. com>
To: math_club@yahoogrou ps.com
Cc: tanvirp123@yahoo. com
Sent: Friday, October 2, 2009 3:43:38 AM
Subject: Re: [math_club] alphabetical math (Ciphered Message)

This assumption can be proven using Goldbach's conjecture but again technically it is not a prove since it depends on Goldbach's conjecture which has not been proven yet.

First note that any odd number can be written as 2(n) +/- 1 where n is even. The reason is as follows:
First any odd number can be written 2(n) - 1; if n is already even then we are done. if n is odd then simply write 2(n-1) + 1 because 2(n) + 1 = 2(n-1) + 1 and now n-1 must be even.

On the other hand Goldbach's conjecture said that any even number can be written as a sum of two prime; so we can write our n as n = p+q where p and q are prime. This shows any odd number can be written as 2(p+q) +/- 1. And this finishes the proof of your conjecture. This is infact stronger than your original assumption since your original assumption is 2(p+/- q) +/- 1.

--- On Thu, 10/1/09, Mirza Sabbir Hossain Beg <mirzasabbirhossainb eg@yahoo. com> wrote:

From: Mirza Sabbir Hossain Beg <mirzasabbirhossainb eg@yahoo. com>
Subject: Re: [math_club] alphabetical math (Ciphered Message)
To: math_club@yahoogrou ps.com
Date: Thursday, October 1, 2009, 11:11 AM

Dear Tanvir, I also gave you another assumption that: any odd numbers can be expressed as this equation- 2(p+/-q)+/-1, where p and q are prime numbers.(stronger than levy's and goldbach's conjecture!! !!!haha.. .ha just kidding man). Anyway, check and verify my assumption that it is unique or not. If it is unique then mail me proof or counter example. Mail me.

From: Tanvir Prince <tanvirp123@yahoo. com>
To: math_club@yahoogrou ps.com
Sent: Wednesday, September 30, 2009 11:10:13 PM
Subject: Re: [math_club] alphabetical math (Ciphered Message)

This conjecture is known as Levy's conjecture. Please read below:

In number theory, Lemoine's conjecture, named after Émile Lemoine, also known as Levy's conjecture, after Hyman Levy, states that all odd integers greater than 5 can be represented as the sum of an odd prime number and an even semiprime. To put it algebraically, 2n + 1 = p + 2q always has a solution in primes p and q (not necessarily distinct) for n > 2. The Lemoine conjecture is similar to but stronger than Goldbach's weak conjecture.
For example, 47 = 13 + 2 × 17 = 37 + 2 × 5 = 41 + 2 × 3 = 43 + 2 × 2. (sequence A046927 in OEIS) counts how many different ways 2n + 1 can be represented as p + 2q.
According to MathWorld, the conjecture has been verified by Corbitt up to 109.
The conjecture was posed by Émile Lemoine in 1895, but in more recent years came to be attributed to Hyman Levy who pondered it in the 1960s.

So as you can see the proof will be extreamly complicated since still it is a open question.

--- On Wed, 9/30/09, Mirza Sabbir Hossain Beg <mirzasabbirhossainb eg@yahoo. com> wrote:

From: Mirza Sabbir Hossain Beg <mirzasabbirhossainb eg@yahoo. com>
Subject: Re: [math_club] alphabetical math (Ciphered Message)
To: math_club@yahoogrou ps.com
Date: Wednesday, September 30, 2009, 3:40 AM

Dear Tanvir, if you think my assumption promising then share it and discuss with all the friends available in this club. Thanks for your mail........

From: Tanvir Prince <tanvirp123@yahoo. com>
To: math_club@yahoogrou ps.com
Sent: Tuesday, September 29, 2009 8:35:48 PM
Subject: Re: [math_club] alphabetical math (Ciphered Message)

Yes. you are absolutely right. I made a mistake. I was not that careful. This assumption looks very promisable to me although I do not have any proof.

--- On Tue, 9/29/09, Mirza Sabbir Hossain Beg <mirzasabbirhossainb eg@yahoo. com> wrote:

From: Mirza Sabbir Hossain Beg <mirzasabbirhossainb eg@yahoo. com>
Subject: Re: [math_club] alphabetical math (Ciphered Message)
To: math_club@yahoogrou ps....com
Date: Tuesday, September 29, 2009, 4:15 AM

Yes Tanvir, you made the mistake because 2*59 + 13 = 131, where both 59 and 13 are prime number. But pls check my assumption for a greater odd number, perhaps there could be a counter example. Above all I like to thank you for your attempt. Mail me if you can prove or disprove it.. Best regards..... ........

From: Tanvir Prince <tanvirp123@yahoo. com>
To: math_club@yahoogrou ps.com
Sent: Tuesday, September 29, 2009 2:31:36 AM
Subject: Re: [math_club] alphabetical math (Ciphered Message)

Your assumption that every odd number bigger than 5 can be written as 2p+q where p and q are prime numbers; I think I found a counter example which is 131.

can you write 131 = 2p + q where p and q prime.
It is possible that I make a mistake.
I use mathematica software to generate a list of numbers of the form 2p+q where p and q runs over all prime numbers starting from 2 to 71 and in that list it looks like 131 is missing..

--- On Mon, 9/28/09, Mirza Sabbir Hossain Beg <mirzasabbirhossainb eg@yahoo. com> wrote:

From: Mirza Sabbir Hossain Beg <mirzasabbirhossainb eg@yahoo. com>
Subject: Re: [math_club] alphabetical math (Ciphered Message)
To: math_club@yahoogrou ps.com
Date: Monday, September 28, 2009, 5:19 AM

Mr. Mamun, my assumption: 1. Any odd numbers greater than 5 can be written as this form: 2p + q, where p and q are prime numbers. 2.... Every prime numbers can be written as this form: 2(p+/-q) +/-1, where p and q are any prime numbers. Send me the proof or counter example if it has so. Thanks...... .......

From: S. M. Mamun Ar Rashid <mamun305@yahoo. com>
To: math_club@yahoogrou ps.com
Sent: Friday, September 25, 2009 10:10:55 PM
Subject: Re: [math_club] alphabetical math (Ciphered Message)

 The problem deals with a ciphered, i.e. encoded message. Your task is to decipher it, that is, to find out a pattern to see whether any meaningful English sentence reveals from it. One approach is to employ the Caesar Method [named after Julius Caesar, who used such technique of exchanging secret and sensitive messages] in which each letter is substituted by another letter. For example, aàc, bàd, càe. If it does not reveal anything, try aàd, bàe, càf. Hope you get the idea. The realm of Secret Message is very ancient, full of conundrums, mysterious, and often bloody.. One story might intrigue you. Mary, the queen of Scots, wanted to assassinate her cousin Queen Elizabeth I of England, and began exchanging secret, ciphered messages with her co-conspirators. But some of Mary's messages were captured by Elizabeth 's spies and after much effort they were cracked by her chief code breaker. Mary was immediately arrested on charge of treason and the deciphered messages were used in trial as evidence. She was found guilty and was beheaded in 1587. All because her cipher was cracked. Regards, Mamun--- On Fri, 9/25/09, mathandsnigdha wrote:From: mathandsnigdha Subject: [math_club] alphabetical mathTo: math_club@yahoogrou ps.comDate: Friday, September 25, 2009, 7:23 AM in "Neurone Anuranan" problem no-116 is disturbing me 4 a long time......i fail to catch it...... plz help me to find out its real meaning..... . ......

• Mr. Mamun, my assumption: every even integer can be written as: 2p - q + 1 and every odd integer as: 2p - q - 2 where p and q are prime number. Is it possible
Message 7 of 16 , Oct 5 6:22 AM
Mr. Mamun, my assumption: every even integer can be written as: 2p - q + 1 and every odd integer as: 2p - q - 2 where p and q are prime number. Is it possible to prove or disprove my assumption? Pls mail me.............

From: S. M. Mamun Ar Rashid <mamun305@...>
To: math_club@yahoogroups.com
Sent: Friday, September 25, 2009 10:10:55 PM
Subject: Re: [math_club] alphabetical math (Ciphered Message)

 The problem deals with a ciphered, i.e. encoded message. Your task is to decipher it, that is, to find out a pattern to see whether any meaningful English sentence reveals from it. One approach is to employ the Caesar Method [named after Julius Caesar, who used such technique of exchanging secret and sensitive messages] in which each letter is substituted by another letter. For example, aàc, bàd, càe. If it does not reveal anything, try aàd, bàe, càf. Hope you get the idea.  The realm of Secret Message is very ancient, full of conundrums, mysterious, and often bloody.. One story might intrigue you. Mary, the queen of Scots, wanted to assassinate her cousin Queen Elizabeth I of England, and began exchanging secret, ciphered messages with her co-conspirators. But some of Mary's messages were captured by Elizabeth 's spies and after much effort they were cracked by her chief code breaker. Mary was immediately arrested on charge of treason and the deciphered messages were used in trial as evidence. She was found guilty and was beheaded in 1587. All because her cipher was cracked.  Regards,  Mamun --- On Fri, 9/25/09, mathandsnigdha wrote:From: mathandsnigdha Subject: [math_club] alphabetical mathTo: math_club@yahoogrou ps.comDate: Friday, September 25, 2009, 7:23 AM in "Neurone Anuranan" problem no-116 is disturbing me 4 a long time......i fail to catch it...... plz help me to find out its real meaning..... . .

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