2019-11-23, 19:46   #326
sweety439

Nov 2016

22×3×211 Posts

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 5-kernel is complete!!!