Problems & Puzzles:
Collection 20th
Coll.20th008.
Primes of the migrant digit
On March 16 2018, Jaime
Ayala wrote:
Find the largest
prime P of length L such that all the integers obtained by migration of
the leftmost digit to each of the righter positions are distinct primes,
being all the primes of the same length L.
Examples
given by Mr. Ayala
L=2,
Primes: 13>31
L=3,
Primes: 149>419 >491
L=6, 984427 > 894427 >
849427 > 844927 > 844297> 844279
Carlos Rivera
found this example and the largest below 2^32:
L=10,
Primes:
{3286760681>2386760681>2836760681>2863760681>2867360681>2867630681>2867603681>2867606381>2867606831>2867606813}
Q. Send your largest one example.
Contributions came from Jan van Delden
***
On July 19, 2018, Jan wrote:
11 digits: 71043402931
12 digits: 352968841549, 712845952529
13 digits: No solution
Admissible numbers:
First digit and last digit in {1,3,7,9}.
Second digit in {1..9}.
Other digits in {0..9}.
All digits, except last one, unequal to the last digit.
Constructing admissible numbers first (with a fixed
length) and testing for prime later has several advantages if the length
becomes large. Sieving as a test for being prime of length L is not an
option anymore. If the length L is larger than 20 the number of
generated admissible numbers is smaller than the number of primes. But
more importantly one could add a test mod 3 and a test mod 11 in the
construction process to determine if any of the resulting migrated
numbers are divisible by 3 or 11, before actually computing these
migrated numbers. For divisor 3 this is just 1 test for all migrated
numbers, but for 11 this test is specific for each migrated number (at
the cost of precomputing L different sums mod 11).
An alternative could be to construct the L migrated numbers in one go. A
disadvantage could be that this would entail unneccesary constructing
numbers that are divisible by 11. So I didn’t pursue this. [If we assume
each migrated number is divisible by 11 with probability p, the
probability of at least 1 migrated number being divisible by 11 would be
1(1p)^L, which tends to 1 with increasing L].
***
