Quote:
Originally Posted by sweety439
b=5, p end with 1: (assume p is a prime in the minimal set of the strings for primes with at least two digits in base 5 other than 10, 12, 21, 23, 32, 34, 43, 111, 131, 133, 313, 401, 414, 14444, 30301, 33001, 33331, 44441, 300031)
* before this 1, p cannot contain 2 (because of 21)
* if before this 1, p contain 1 > assume p is {xxx}1{yyy}1 > y cannot contain 0, 1, 2, or 3 (because of 10, 111, 12, and 131)
** if y is not empty > y contain only the digits 4 > x cannot contain 1, 2, 3, or 4 (because of 111, 21, 34, and 414) > x contain only the digits 0 (a contradiction, since a number cannot have leading zeros)
** thus y is empty > p is {xxx}11 > x cannot contain 1 or 2 (because of 111 and 21), but x cannot contain both 3 and 4 (because of 34 and 43) > x contain either only 0 and 3, or only 0 and 4, however, {0,3}11 is divisible by 3, and {0,4}11 is divisible by 2, and neither can be primes, a contradiction!!!
> thus, before this 1, p cannot contain 1

Therefore, before this 1, p can only contain the digits 0, 3, and 4.
However, p cannot contain both 3 and 4 (because of 34 and 43)
thus, before this 1, p contain either only 0 and 3, or only 0 and 4
For the primes contain only 0 and 3:
Since the first digit cannot be 0, it can only be 3
thus p is 3{xxx}1
x should contain at least one 3 (or p is of the form 3{0}1, and 3{0}1 is divisible by 2)
thus we can assume p is 3{xxx}3{yyy}1
* if both x and y contain 0, then we have the prime 30301
* if x does not contain 0, then x contain only 3
** if x contain at least two 3, then we have the prime 33331
** if x contain exactly one 3, then p is 3{0}3{0}3{yyy}1, and y can contain only the digits 0 (y cannot contain 3, because of 33331) > p is of the form 3{0}3{0}3{0}1 and divisible by 2, a contradiction
** if x is empty, then p is 33{yyy}1, and y can contain only the digits 0 (y cannot contain at least two 3 because of 33331, and if y contain exactly one 3, then p is divisible by 2 and cannot be prime), then we have the prime 33001
* if y does not contain 0, then y contain only 3
** if y contain at least two 3, then we have the prime 33331
** if y contain exactly one 3, then p is 3{xxx}3{0}3{0}1, and x can contain only the digits 0 (x cannot contain 3, because of 33331) > p is of the form 3{0}3{0}3{0}1 and divisible by 2, a contradiction
** if y is empty, then p is 3{xxx}31, and x can contain only the digits 0 (x cannot contain at least two 3 because of 33331, and if x contain exactly one 3, then p is divisible by 2 and cannot be prime), then we have the prime 300031
For the primes contain only 0 and 4:
Since the first digit cannot be 0, it can only be 4
thus p is 4{xxx}1
if x contain 0, then we have the prime 401
if x does not contain 0 (thus contain only the digit 4), then we have the prime 44441
Therefore, no such prime p can exist!!!
The 5kernel is complete!!!