20041023, 06:39  #1 
Oct 2004
3 Posts 
Smallest tenmilliondigit prime
Eventually we will want to find the smallest tenmilliondigit prime.
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 
20041023, 07:50  #2 
Nov 2002
2·37 Posts 
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

20041023, 11:42  #3  
Banned
"Luigi"
Aug 2002
Team Italia
2^{2}·1,193 Posts 
Quote:
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. Luigi 

20041025, 13:15  #4  
Oct 2004
Austria
7·353 Posts 
Quote:
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. 

20041027, 19:52  #5 
Oct 2004
Austria
7·353 Posts 
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. 
20041028, 10:54  #6 
Oct 2004
Austria
7·353 Posts 
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 :this thread) 7  found by heck 13  can be read out of hecks table 11  "manual" factoring using divisibility rule for p=11 
20041028, 11:18  #7  
Nov 2003
2^{6}·113 Posts 
Quote:
algebra might observe that if a is odd, then 10^(ab) + 1 is divisible by 10^b+1. What is it that compels people to blindly throw a calculator or computer at a problem *BEFORE* doing any thinking about the mathematics involved??? 

20041028, 11:18  #8 
Oct 2004
Austria
4647_{8} Posts 
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. 
20041028, 11:32  #9  
Oct 2004
Austria
9A7_{16} Posts 
Quote:
I figured out 11 and 13 *without* using a calculator by looking at heck's table in this thread 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 Last fiddled with by Andi47 on 20041028 at 11:39 Reason: Just another bit of information 

20041028, 11:34  #10  
"Sander"
Oct 2002
52.345322,5.52471
29×41 Posts 
Quote:
Last fiddled with by smh on 20041028 at 11:40 

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