Smallest tenmilliondigit prime
Eventually we will want to find the [I]smallest[/I] tenmilliondigit prime.:smile:
mighty big number, factor 10^9999999 + 1, 7 10^9999999 + 3, 23 10^9999999 + 7, 11852969 10^9999999 + 9, 47 10^9999999 + 11, 3 10^9999999 + 13, 277 10^9999999 + 17, 3 10^9999999 + 19, 7583 10^9999999 + 21, 718453 10^9999999 + 23, 3 10^9999999 + 27, 13 10^9999999 + 29, 3 10^9999999 + 31, 821879 etc.. *Heck :cat: 
but this will only help to tell you that the numbers are NOT PRIME because the primilty tests for such big numbers ( without a suitable form like Mersenne Numbers) will take some hundred years of computing to verify that it is prime

[QUOTE=andi314]but this will only help to tell you that the numbers are NOT PRIME because the primilty tests for such big numbers ( without a suitable form like Mersenne Numbers) will take some hundred years of computing to verify that it is prime[/QUOTE]
It is partially what we are doing for Mersennes at LMH > 79.2M. Maybe it is useful to have a table of factored numbers from where someone else will start checking primality. My hint: avoiding sums wirh numbers of the form 3n+2, as they will always have 3 as a factor will spped up search. :wink: Luigi 
[QUOTE=Heck]Eventually we will want to find the [I]smallest[/I] tenmilliondigit prime.:smile:
mighty big number, factor 10^9999999 + 1, 7 10^9999999 + 3, 23 10^9999999 + 7, 11852969 10^9999999 + 9, 47 10^9999999 + 11, 3 10^9999999 + 13, 277 10^9999999 + 17, 3 10^9999999 + 19, 7583 10^9999999 + 21, 718453 10^9999999 + 23, 3 10^9999999 + 27, 13 10^9999999 + 29, 3 10^9999999 + 31, 821879 etc.. [/QUOTE] This gives us the following moduloresults, if somebody wants to start sieving: let a be 10^9999999; let b be 1, 3, 7, ... a+1 ≡ 2 mod 3 a+1 ≡ 1 mod 5 (obviously) a+1 ≡ 0 mod 7 a+1 ≡ 0 mod 11 (each number of the form 1000000....00001 which has an even number of digits is divisible by 11; (a+1)/11 = 90909090.......09091; 9999998 digits) a+1 ≡ 0 mod 13 a+1 ≡ ??? mod 17 a+1 ≡ ??? mod 19 a+1 ≡ 21 mod 23 a+1 ≡ ??? mod 29, 31, 37, 41, 43 a+1 ≡ 39 mod 47 It gives us also two more factors of 10^9999999+1, namely 11 and 13. :banana: 
1 Attachment(s)
I started some sieving, the sieve data for 10^9999999+1 to 10^9999999+200k is attatched to this posting.
I sieved out numbers divisible by the primes p = 3, 5, 7, 11 and 13. Heck, if You can provide some more small divisors of this huge numbers, especially 17, 19, ..., I will continue sieving. 
I just did some trial division on 10^9999999+1  the lowest odd 10 million digit number and found another small factor: 19 divides 10^9999999+1.
Other factors found so far (see also :[URL=http://www.mersenneforum.org/showthread.php?t=3227]this thread[/URL]) 7  found by heck 13  can be read out of hecks table 11  "manual" factoring using divisibility rule for p=11 
[QUOTE=Andreas Schinde]I just did some trial division on 10^9999999+1  the lowest odd 10 million digit number and found another small factor: 19 divides 10^9999999+1.
Other factors found so far (see also :[URL=http://www.mersenneforum.org/showthread.php?t=3227]this thread[/URL]) 7  found by heck 13  can be read out of hecks table 11  "manual" factoring using divisibility rule for p=11[/QUOTE] Some intelligent, or with an understanding of 8th grade level first year algebra might observe that if a is odd, then 10^(ab) + 1 is divisible by 10^b+1. :cry: What is it that compels people to blindly throw a calculator or computer at a problem *BEFORE* doing any thinking about the mathematics involved??? :glare: 
1 Attachment(s)
I did some more trial factoring on 10^9999999+1, no more divisor up to 10000.
The attatched file lists the modulo residues 10^9999999+1 == c mod p for p = 7 to 10k. 
[QUOTE=Bob Silverman]Some intelligent, or with an understanding of 8th grade level first year
algebra might observe that if a is odd, then 10^(ab) + 1 is divisible by 10^b+1. :cry: What is it that compels people to blindly throw a calculator or computer at a problem *BEFORE* doing any thinking about the mathematics involved??? :glare:[/QUOTE] I dont't know how Heck figured out the divisor 7 (maybe it's this bit of algebra you mentioned with b=1001 which gives the factors 7, 11 and 13) I figured out 11 and 13 *without* using a calculator by looking at heck's table in [URL=http://www.mersenneforum.org/showthread.php?t=3227]this thread[/URL] and by rule that 10^a+1 is divisible by 11 if a is odd. Edit: Just to mention it: divisibility by 10^b+1 gives us the factors 10^3+1 = 7*11*13 10^9+1 = 7*11*13*19*52579 and so on for b = 239, 4649 and 239*3, 239*9, 239*4649, 239*4649*3, 4649*3, 4649*9 :ernst: 
[QUOTE=Andreas Schinde]Other factors found so far (see also :[URL=http://www.mersenneforum.org/showthread.php?t=3227]this thread[/URL])[/quote]
Moved things to this thread. Had nothing to do with the subject of the other. 
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