Re: Re: Finding Primes

Any two 200 digit primes will do it

Without knowing all the 200 digit primes, I can't say whether or not that is a true statement. I wouldn't assume it is sufficient, however. The square root of 10^399 is a little greater than this 200 digit number:

3162277660168379331998893544432718533719555139325216826857504852792594
+438639238221344248108379300295187347284152840055148548856030453880014
+6905195967001539033449216571792599406591501534741133394841240
[download]

So, if there are two primes between 10^199 and that number, then there will be two 200 digit primes whose product is not a 400 digit number.

-sauoq
"My two cents aren't worth a dime.";

Comment on Re: Re: Finding Primes Download Code

Replies are listed 'Best First'.
Re: Re: Re: Finding Primes by Anonymous Monk on Aug 14, 2003 at 09:36 UTC
First of all, there are an infinite number of primes and there is a theorem by Bertrand which states that there is a prime between any natural number n and 2*n. So there are two prime numbers which product is a 400 digit number.	[reply]
Re: Re: Re: Re: Finding Primes by sauoq (Abbot) on Aug 14, 2003 at 14:10 UTC
Very good, but why are you replying to me? I was arguing that the product of two 200 digit primes is not necessarily 400 digits. It may only be 399. Perhaps you'd like to tell the OP that he can prove the existence of such primes without finding them. I expect that was what his professor had in mind anyway... -sauoq "My two cents aren't worth a dime.";	[reply]
Re: Re: Re: Finding Primes by thor (Priest) on Aug 14, 2003 at 12:37 UTC
Well, as 10^1 is a 2 digit number, 10^399 is a 400 digit number....:) thor	[reply]
A reply falls below the community's threshold of quality. You may see it by logging in.